MHB Arithmetic for Quotient Groups - How exaclty does it work

[Math Amateur]

I have just received some help from Euge regarding the proof of part of the Correspondence Theorem (Lattice Isomorphism Theorem) for groups ...

But Euge has made me realize that I do not understand quotient groups well enough ... here is the issue coming from Euge's post ...

We are to consider the inclusions

$$10 \mathbb{Z} \subseteq 5 \mathbb{Z} \subseteq \mathbb{Z} $$

We are then asked to "factor each of these groups out by $$10 \mathbb{Z}$$My question is, how EXACTLY do we carry out the arithmetic involved and why?
Some thoughts ... ...
We have that ...

$$\mathbb{Z} = \{ \ ... \ ... \ -3, -2, -1, 0, 1, 2, 3, 4, 5, \ ... \ ... \} $$$$5 \mathbb{Z} = \{ \ ... \ ... \ -10, -5, 0 , 5, 10, \ ... \ ... \}$$$$10 \mathbb{Z} = \{ \ ... \ ... \ -20, -10, 0, 10, 20 \ ... \ ... \}$$
Now, how exactly do we form $$\mathbb{Z} / 10 \mathbb{Z}$$ , $$ 5 \mathbb{Z}/ 10 \mathbb{Z}$$ and $$10 \mathbb{Z} / 10 \mathbb{Z}$$ ... what is the process ... what is the exact arithmetic to make this happen ...?
I am aware, or I think that $$ \mathbb{Z} / 10 \mathbb{Z} = \{ \overline{0}, \overline{1}, \overline{2}, \ ... \ ... \ \overline{9} \} $$but how did I actually get there ... ... not completely sure ... and that is a problem ...Similarly for the other factorings ...I think that $$5 \mathbb{Z} / 10 \mathbb{Z} = \{ \overline{0}, \overline{1} \}
$$and $$10 \mathbb{Z} / 10 \mathbb{Z} = \{ \overline{0} \} $$But what is the actual arithmetic process and what is the meaning of what is going on...?Hope someone can provide a clear picture of the actual process taking place and the meaning of what is happening ...

Peter

 

[Deveno]

Well, with groups, we start with a normal subgroup. The normality is *crucial*, otherwise the "arithmetic" on the cosets isn't well-defined.

Abelian groups are conceptually easier to deal with, since normality is automatic. In this case, *any* subgroup will do.

The first step is to partition $G$ via the cosets of our (normal) subgroup $H$. Every coset has several members (unless we partition by the trivial subgroup) and its members are called "representatives". The basic idea is we perform the basic operation on the representatives, and then "forget" which particular representatives we used.

Let's look at $\Bbb Z/5\Bbb Z$, since you brought it up. Note our partition of $\Bbb Z$ is THIS one:

$\Bbb Z = \{\dots, -15, -10,-5,0,5,10,15,\dots\} \cup \{\dots,-14,-9,-4,1,6,11,16,\dots\} \cup \{\dots,-13,-8,-3,2,7,12,17,\dots\} \cup \{\dots,-12,-7,-2,3,8,13,18,\dots\} \cup \{\dots,-11,-6,-1,4,9,14,19,\dots\}$

For the moment, let's just call these $A,B,C,D,E$ and $F$.

So how do we find, for example, $B+C$? Well, we take any $b \in B$, and any $c \in C$. Then we compute $b+c$ in the "main group" $\Bbb Z$. Let's do this: we have $-9 \in B$, and $7 \in C$. So $b + c = -2$ in this case. Which coset is $-2$ in? $D$, so we have:

$B + C = D$.

But wait, you say, how do we know this will "always work out"? What if this was just a coincidence, because this formed some "special example"? Well, in this particular case (a quotient group of the integers), we have some "canonical" representatives, the first $5$ non-negative integers: $0,1,2,3,4$. Because of the division algorithm (a special feature of the integers), we can write, for any integer $n$:

$n = 5q + r$, where $r \in \{0,1,2,3,4\}$. Note each of these lie in distinct cosets of $5\Bbb Z$ (our partition above).

If we write $n' = 5q' + r'$ (the same way), we have:

$n + n' = 5q + r + 5q' + r' = 5(q+q') + r+r'$. Let's break this down into cases:

$r + r' < 5$ -in this case, we can just use $r + r'$ as our representative for $n + n'$.

$r + r' = 5$ -in this case, we have $n + n' = 5(q + q') + r + r' = 5(q + q' + 1) + 0$, and we can use $0$ as our representative for $n + n'$.

$r + r' = 5 + t$, where $t \in \{1,2,3,4\}$ (this is as big as we can get, since each of $r,r' < 5$). Here, we have:

$n + n' = 5(q + q') + r + r' = 5(q + q') + 5 + t = 5(q + q' + 1) + t$, and $t$ can stand for $n + n'$.

You can see that this gives us a way to "resolve" $n + n'$ mod $5$, but how do we know this works *consistently*?

Let's phrase it another way:

If $n = 5q + r$, then we can see $m$ lies in the same coset as $n$ if and only if $m = 5q' + r$ (the same $r$, perhaps a different $q'$).

In particular, $m - n = 5q' + r - 5q - r = 5q' - 5q = 5(q' - q)$, that is, the DIFFERENCE of $m$ and $n$ lies in $5\Bbb Z$ (note that since $-5\Bbb Z = 5\Bbb Z$ we could have used $n - m$ just as easily-subgroups are closed under inversion). This turns out to be something we can generalize to arbitrary groups. We'll get back to this is a moment.

We say that $m \equiv n$ modulo $5$ if $m - n \in 5\Bbb Z$. It is straight-forward to see this defines an equivalence relation on $\Bbb Z$:

$m \equiv m$ modulo $5$, since $m - m = 0 = 5\cdot 0 \in 5\Bbb Z$

$m \equiv n$ modulo $5$ implies $n \equiv m$ modulo $5$ (we discussed this above).

Suppose $m \equiv n$ modulo $5$, and $n \equiv k$ modulo $5$.

Then $m - k = (m - n) + (n - k) = 5a + 5b = 5(a + b)$, so $m \equiv k$ modulo $5$ (this is because $5\Bbb Z$ is closed under addition).

So let's call the equivalence class of $n$, under this equivalence, $\overline{n}$. We'd like to show that we can DEFINE:

$\overline{m} + \overline{n} = \overline{m+n}$.

To do this, we need to show it doesn't matter which "$m$" we choose of of the many possibilities (members) that $\overline{m}$ contains, and similarly for $\overline{n}$. Explicitly, we need to demonstrate that if:

$\overline{m} = \overline{m'}$ and $\overline{n} = \overline{n'}$ that:

$\overline{m+n} = \overline{m'+n'}$, so that we get a "well-defined result".

Now if $\overline{m} = \overline{m'}$ , then $m - m' = 5s$, for some integer $s$. Similarly:

$\overline{n} = \overline{n'}$ means $n - n' = 5t$, for some integer $t$.

Thus $m+n - (m' + n') = m + n - m' - n' = (m - m') + (n - n') = 5s + 5t = 5(s + t)$, and $m+n \equiv m'+n'$ modulo $5$. Note how we used commutativity of addition in $\Bbb Z$ to re-arrange the terms.

So, to compute $B + C = \overline{1} + \overline{2}$, we find we are looking for $\overline{1 + 2} = \overline{3}$, which is $D$.

We can generalize this, now, to an arbitrary group $G$, and a subgroup $H$. We say that $a,b \in G$ are equivalent modulo $H$ if:

$ab^{-1} \in H$ (since $H$ is a subgroup, this is equivalent to $(ab^{-1})^{-1} = (b^{-1})^{-1}a^{-1} = ba^{-1} \in H$, as well).

This ALWAYS defines a partition of $G$ into cosets of $H$. Let's use left cosets, for now. So we can write:

$G = \bigcup\limits_i g_iH$, as a disjoint union.

This union consists of $H$-sized clumps of $G$. If $G$ is finite, we get a finite number of "clumps", and there are FEWER of these, so we have a SMALLER set of "clumps".

The question is, can we multiply these together, somehow?

What we would like, is for $(aH)(bH) = abH$.

There is a problem though: the set on the left is all elements of $G$ of the form:

$\{ah_1bh_2: h_1,h_2 \in H\}$

There is no reason to suppose, in general, that $ah_1bh_2 = abh_3$, for some $h_3 \in H$.

If, however, we had, for ANY $b \in G$, that $h_1b = bh'$, for any $h_1 \in H$, and some $h'$ (depending on $h_1$), we COULD write:

$ah_1bh_2 = a(h_1b)h_2 = a(bh')h_2 = (ab)h'h_2$, and take $h_3 = h'h_2$.

In other words, if $bH = Hb$, for any $b \in G$, that is precisely what we need of $H$ for our generalization to work.

Such subgroups are called NORMAL, and they are the "special" ones we are after. That let's us do the same "term re-arrangement" that worked so well for the integers modulo $5$.

I'm out of time, for now, so I'll be back later for "quotients of quotients".

 

[Math Amateur]

Well, with groups, we start with a normal subgroup. The normality is *crucial*, otherwise the "arithmetic" on the cosets isn't well-defined.

Abelian groups are conceptually easier to deal with, since normality is automatic. In this case, *any* subgroup will do.

The first step is to partition $G$ via the cosets of our (normal) subgroup $H$. Every coset has several members (unless we partition by the trivial subgroup) and its members are called "representatives". The basic idea is we perform the basic operation on the representatives, and then "forget" which particular representatives we used.

Let's look at $\Bbb Z/5\Bbb Z$, since you brought it up. Note our partition of $\Bbb Z$ is THIS one:

$\Bbb Z = \{\dots, -15, -10,-5,0,5,10,15,\dots\} \cup \{\dots,-14,-9,-4,1,6,11,16,\dots\} \cup \{\dots,-13,-8,-3,2,7,12,17,\dots\} \cup \{\dots,-12,-7,-2,3,8,13,18,\dots\} \cup \{\dots,-11,-6,-1,4,9,14,19,\dots\}$

For the moment, let's just call these $A,B,C,D,E$ and $F$.

So how do we find, for example, $B+C$? Well, we take any $b \in B$, and any $c \in C$. Then we compute $b+c$ in the "main group" $\Bbb Z$. Let's do this: we have $-9 \in B$, and $7 \in C$. So $b + c = -2$ in this case. Which coset is $-2$ in? $D$, so we have:

$B + C = D$.

But wait, you say, how do we know this will "always work out"? What if this was just a coincidence, because this formed some "special example"? Well, in this particular case (a quotient group of the integers), we have some "canonical" representatives, the first $5$ non-negative integers: $0,1,2,3,4$. Because of the division algorithm (a special feature of the integers), we can write, for any integer $n$:

$n = 5q + r$, where $r \in \{0,1,2,3,4\}$. Note each of these lie in distinct cosets of $5\Bbb Z$ (our partition above).

If we write $n' = 5q' + r'$ (the same way), we have:

$n + n' = 5q + r + 5q' + r' = 5(q+q') + r+r'$. Let's break this down into cases:

$r + r' < 5$ -in this case, we can just use $r + r'$ as our representative for $n + n'$.

$r + r' = 5$ -in this case, we have $n + n' = 5(q + q') + r + r' = 5(q + q' + 1) + 0$, and we can use $0$ as our representative for $n + n'$.

$r + r' = 5 + t$, where $t \in \{1,2,3,4\}$ (this is as big as we can get, since each of $r,r' < 5$). Here, we have:

$n + n' = 5(q + q') + r + r' = 5(q + q') + 5 + t = 5(q + q' + 1) + t$, and $t$ can stand for $n + n'$.

You can see that this gives us a way to "resolve" $n + n'$ mod $5$, but how do we know this works *consistently*?

Let's phrase it another way:

If $n = 5q + r$, then we can see $m$ lies in the same coset as $n$ if and only if $m = 5q' + r$ (the same $r$, perhaps a different $q'$).

In particular, $m - n = 5q' + r - 5q - r = 5q' - 5q = 5(q' - q)$, that is, the DIFFERENCE of $m$ and $n$ lies in $5\Bbb Z$ (note that since $-5\Bbb Z = 5\Bbb Z$ we could have used $n - m$ just as easily-subgroups are closed under inversion). This turns out to be something we can generalize to arbitrary groups. We'll get back to this is a moment.

We say that $m \equiv n$ modulo $5$ if $m - n \in 5\Bbb Z$. It is straight-forward to see this defines an equivalence relation on $\Bbb Z$:

$m \equiv m$ modulo $5$, since $m - m = 0 = 5\cdot 0 \in 5\Bbb Z$

$m \equiv n$ modulo $5$ implies $n \equiv m$ modulo $5$ (we discussed this above).

Suppose $m \equiv n$ modulo $5$, and $n \equiv k$ modulo $5$.

Then $m - k = (m - n) + (n - k) = 5a + 5b = 5(a + b)$, so $m \equiv k$ modulo $5$ (this is because $5\Bbb Z$ is closed under addition).

So let's call the equivalence class of $n$, under this equivalence, $\overline{n}$. We'd like to show that we can DEFINE:

$\overline{m} + \overline{n} = \overline{m+n}$.

To do this, we need to show it doesn't matter which "$m$" we choose of of the many possibilities (members) that $\overline{m}$ contains, and similarly for $\overline{n}$. Explicitly, we need to demonstrate that if:

$\overline{m} = \overline{m'}$ and $\overline{n} = \overline{n'}$ that:

$\overline{m+n} = \overline{m'+n'}$, so that we get a "well-defined result".

Now if $\overline{m} = \overline{m'}$ , then $m - m' = 5s$, for some integer $s$. Similarly:

$\overline{n} = \overline{n'}$ means $n - n' = 5t$, for some integer $t$.

Thus $m+n - (m' + n') = m + n - m' - n' = (m - m') + (n - n') = 5s + 5t = 5(s + t)$, and $m+n \equiv m'+n'$ modulo $5$. Note how we used commutativity of addition in $\Bbb Z$ to re-arrange the terms.

So, to compute $B + C = \overline{1} + \overline{2}$, we find we are looking for $\overline{1 + 2} = \overline{3}$, which is $D$.

We can generalize this, now, to an arbitrary group $G$, and a subgroup $H$. We say that $a,b \in G$ are equivalent modulo $H$ if:

$ab^{-1} \in H$ (since $H$ is a subgroup, this is equivalent to $(ab^{-1})^{-1} = (b^{-1})^{-1}a^{-1} = ba^{-1} \in H$, as well).

This ALWAYS defines a partition of $G$ into cosets of $H$. Let's use left cosets, for now. So we can write:

$G = \bigcup\limits_i g_iH$, as a disjoint union.

This union consists of $H$-sized clumps of $G$. If $G$ is finite, we get a finite number of "clumps", and there are FEWER of these, so we have a SMALLER set of "clumps".

The question is, can we multiply these together, somehow?

What we would like, is for $(aH)(bH) = abH$.

There is a problem though: the set on the left is all elements of $G$ of the form:

$\{ah_1bh_2: h_1,h_2 \in H\}$

There is no reason to suppose, in general, that $ah_1bh_2 = abh_3$, for some $h_3 \in H$.

If, however, we had, for ANY $b \in G$, that $h_1b = bh'$, for any $h_1 \in H$, and some $h'$ (depending on $h_1$), we COULD write:

$ah_1bh_2 = a(h_1b)h_2 = a(bh')h_2 = (ab)h'h_2$, and take $h_3 = h'h_2$.

In other words, if $bH = Hb$, for any $b \in G$, that is precisely what we need of $H$ for our generalization to work.

Such subgroups are called NORMAL, and they are the "special" ones we are after. That let's us do the same "term re-arrangement" that worked so well for the integers modulo $5$.

I'm out of time, for now, so I'll be back later for "quotients of quotients".

Thanks for the help and support Deveno ... looks really great at first sight ...

... will now now work carefully through the details of your post ...

Thanks again ...

Peter

 

[Math Amateur]

Thanks for the help and support Deveno ... looks really great at first sight ...

... will now now work carefully through the details of your post ...

Thanks again ...

Peter

Hi Deveno,

Worked through you post in detail ... all was very clear ... extremely helpful ...

So, I think I have a better understanding of quotient groups ...

But still very unsure about the exact process for the situation where one is given $$5 \mathbb{Z}$$ and $$10 \mathbb{Z}$$ and one wants to form $$5 \mathbb{Z} / 10 \mathbb{Z}$$ and $$10 \mathbb{Z} / 10 \mathbb{Z}$$ ... ...

Hope you can help to clarify this situation as well ...

Thanks again for your helpful post ...

Peter

 

[Deveno]

OK, so where was I?

Basically, the process of "quotienting" (is that even a word?) is "identifying" things that aren't "equal". This is the entire logic behind an equivalence relation. Now if an algebraic operation on a set $S$ respects an equivalence on $S$, said equivalence is called a congruence on $S$ with respect to our operation.

For groups, this means our "clumps" (cosets) when multiplied "don't cross clump borders": if $a$ is in coset $A$, and $b$ is coset $B$, then for any $x \in A$ and $y \in B$, $xy$ is in the same coset (clump) as $ab$ is.

This means any such $x$ and $y$ (it doesn't MATTER which "representatives" we choose) can be used to calculate $[ab]$ (that is the coset $abH$, when our subgroup determining the equivalence is $H$). Let's look at a specific group, and two possible subgroups to see how this works, and how it doesn't if we pick "the wrong kind of subgroup".

So let $G = \{e,a,a^2,b,ab,a^2b\}$, where $a^3 = b^2 = e$ and $ba = a^2b$. This is (the only, up to isomorphism) a non-abelian group of order $6$.

Two obvious subgroups are $N = \{e,a,a^2\}$ and $H = \{e,b\}$. Let's look at the set of left cosets for each. We'll start with $N$.

There are just two (left) cosets, $N$, and $bN = \{b,ba,ba^2\}$. Now we know (by definition) that $ba = a^2b$. It seems logical that $ba^2$ must be the only element of $G$ left, $ab$. We can prove this, just from the definition:

$ba^2 = (ba)a = (a^2b)a = a^2(ba) = a^2(a^2b) = a^4b = (a^3a)b = (ea)b = ab$.

If this is to be a group, we must have (as with ANY group of order $2$):

$NN = N$
$N(bN) = bN$
$(bN)N = bN$
$(bN)(bN) = N$.

The first is clear, since $N$ is a subgroup, by closure. Now since $e \in N$, we have:

$bN = (eb)N = e(bN) \subseteq N(bN)$.

So to show $N(bN) \subseteq bN$ we need only verify that $a(bN) \subseteq bN$ and $a^2(bN) \subseteq bN$.

In fact: $a(bN) = \{ab,a(ab),a(a^2b)\} = \{ab,a^2b,b\} = bN$, and it is easily seen that $a^2(bN) = bN$, also. I leave it to you to show that $(bN)N = bN$ (use the fact that we have $ba = a^2b$ and $ba^2 = ab$).

The last of the $4$ equations is more interesting: we have $9$ products to check:

$bb = b^2 = e \in N$
$b(ab) = b(ba^2) = b^2a^2 = ea^2 = a^2 \in N$
$b(a^2b) = b(ba) = b^2a = a \in N$
$(ab)b = ab^2 = a \in N$
$(ab)(ab) = a(ba)b = a(a^2b)b = a^3b^2 = ee = e \in N$
$(ab)(a^2b) = a(ba)ab = a(a^2b)ab = a^3(bab) = bab = (ba)b = (a^2b)b = a^2b^2 = a^2 \in N$
$(a^2b)b = a^2b^2 = a^2 \in N$
$(a^2b)(ab) = a^2(ba)b = a^2(a^2b)b = a^4b^2 = a \in N$
$(a^2b)(a^2b) = a^2(ba)ab = a^2(a^2b)ab = a^4(bab) = a(bab) = (ab)(ab) = e \in N$.

So, indeed the multiplication does "work out", and in fact, we can see that:

$gN = Ng$ for any $g \in G$ (this is obvious for $g \in N$, perhaps you can see why it works out for $g \in bN$). Indeed, arguing from the fact that "mod $N$" is an equivalence relation, we can see that $G$ has just two cosets, and since the left and right cosets of $eN = N = Ne$ coincide, the "only other coset" must coincide (left and right) for $g \not\in N$.

By contrast, let's look at the left cosets of $H$:

$H = \{e,b\}$
$aH = \{a,ab\}$
$a^2H = \{a^2,a^2b\}$.

If we are to have a group, we must have (since our group is of order $3$, the two non-identity elements must be inverses):

$(aH)(a^2H) = H$.

Again, we look at all $4$ possible products:

$a(a^2) = a^3 = e \in H$
$a(a^2b) = a^3b = b \in H$. So far, so good.
$(ab)(a^2) = a(ba)a = a(a^2b)a = a^3(ba) = ba = a^2b$. Huh. This ISN'T in $H$.
$(ab)(a^2b) = a(ba)ab = a(a^2b)ab = a^3(bab) = bab = a^2b^2 = a^2$. Neither is this.

The problem here is that $aH = \{a,ab\}$, but $Ha = \{a,ba\} = \{a,a^2b\} \neq aH$.

So we can't make the set of left cosets of $H$ into a group, it just doesn't work. The equivalence doesn't respect the group operation.

Now all of this is "at the element level", we're looking at the "nitty gritty" details of individual products. Let's look at the situation "from the outside".

With $N$, which happens to be normal, we can define in a perfectly acceptable way:

$(xN)(yN) = (xy)N$.

The above equation means that $x \mapsto [x]_{\sim_N} = xN$ is a homomorphism.

What does this homomorphism DO? For our $G$ and $N$, it basically says "let's turn all the $a$'s into the identity". We can see that this approach DOESN'T work for the $b$'s because it MATTERS which side (front or back) an $a$ is on with respect to $b$:

$ba = a^2b$ and $ab = ba^2$

So "turning $b$'s to $e$" would give $a = a^2$, which is not how things ARE in $G$.

But "turning $a$'s to $e$" DOES work, because it leads to $b = b$ (or $e = e$, if we just have $a$'s), which is fine.

So let's look at this situation just from knowing we have a congruence (normal subgroup induced equivalence), but don't know "the element details". We get a homomorphism:

$G \to G/N$ in which $G$ gets "condensed" ($N$ is "factored" out).

In this light, $\Bbb Z/5\Bbb Z$ is "identifying" $5$ with $0$, so we only get $5$ "virtually different" elements:

0 is 5 is 10 is...
1 is 6 is 11 is..
2 is 7 is 12...

and so on. All the "other" properties of the integers (commutativity, being cyclic) get "kept" we just have "five-ness" and 4 other kinds of "non-five-ness".

EVERY QUOTIENT GROUP ARISES IN THIS WAY, from an equivalence induced by a homomorphism. The equivalence is:

$g \sim g'$ if $\phi(g) = \phi(g')$ for our homomorphism $\phi:G \to \phi(G)$.

The normal subgroup we "quotient by" is $\text{ker }\phi = \phi^{-1}(e)$. It is straight-forward to show this is indeed a normal subgroup of $G$, and the existence of the (canonical) homomorphism $G \to G/N$ for any normal subgroup $N$ shows any normal subgroup gives rise to the kernel of a homomorphism.

I'll say this again, for emphasis: normal subgroups are kernels of homomorphisms.

Now, we are no longer talking about "elements", but just special kinds of mappings between groups. What kind of mappings? Surjective homomorphisms. More in post #2.

 

[Math Amateur]

OK, so where was I?

Basically, the process of "quotienting" (is that even a word?) is "identifying" things that aren't "equal". This is the entire logic behind an equivalence relation. Now if an algebraic operation on a set $S$ respects an equivalence on $S$, said equivalence is called a congruence on $S$ with respect to our operation.

For groups, this means our "clumps" (cosets) when multiplied "don't cross clump borders": if $a$ is in coset $A$, and $b$ is coset $B$, then for any $x \in A$ and $y \in B$, $xy$ is in the same coset (clump) as $ab$ is.

This means any such $x$ and $y$ (it doesn't MATTER which "representatives" we choose) can be used to calculate $[ab]$ (that is the coset $abH$, when our subgroup determining the equivalence is $H$). Let's look at a specific group, and two possible subgroups to see how this works, and how it doesn't if we pick "the wrong kind of subgroup".

So let $G = \{e,a,a^2,b,ab,a^2b\}$, where $a^3 = b^2 = e$ and $ba = a^2b$. This is (the only, up to isomorphism) a non-abelian group of order $6$.

Two obvious subgroups are $N = \{e,a,a^2\}$ and $H = \{e,b\}$. Let's look at the set of left cosets for each. We'll start with $N$.

There are just two (left) cosets, $N$, and $bN = \{b,ba,ba^2\}$. Now we know (by definition) that $ba = a^2b$. It seems logical that $ba^2$ must be the only element of $G$ left, $ab$. We can prove this, just from the definition:

$ba^2 = (ba)a = (a^2b)a = a^2(ba) = a^2(a^2b) = a^4b = (a^3a)b = (ea)b = ab$.

If this is to be a group, we must have (as with ANY group of order $2$):

$NN = N$
$N(bN) = bN$
$(bN)N = bN$
$(bN)(bN) = N$.

The first is clear, since $N$ is a subgroup, by closure. Now since $e \in N$, we have:

$bN = (eb)N = e(bN) \subseteq N(bN)$.

So to show $N(bN) \subseteq bN$ we need only verify that $a(bN) \subseteq bN$ and $a^2(bN) \subseteq bN$.

In fact: $a(bN) = \{ab,a(ab),a(a^2b)\} = \{ab,a^2b,b\} = bN$, and it is easily seen that $a^2(bN) = bN$, also. I leave it to you to show that $(bN)N = bN$ (use the fact that we have $ba = a^2b$ and $ba^2 = ab$).

The last of the $4$ equations is more interesting: we have $9$ products to check:

$bb = b^2 = e \in N$
$b(ab) = b(ba^2) = b^2a^2 = ea^2 = a^2 \in N$
$b(a^2b) = b(ba) = b^2a = a \in N$
$(ab)b = ab^2 = a \in N$
$(ab)(ab) = a(ba)b = a(a^2b)b = a^3b^2 = ee = e \in N$
$(ab)(a^2b) = a(ba)ab = a(a^2b)ab = a^3(bab) = bab = (ba)b = (a^2b)b = a^2b^2 = a^2 \in N$
$(a^2b)b = a^2b^2 = a^2 \in N$
$(a^2b)(ab) = a^2(ba)b = a^2(a^2b)b = a^4b^2 = a \in N$
$(a^2b)(a^2b) = a^2(ba)ab = a^2(a^2b)ab = a^4(bab) = a(bab) = (ab)(ab) = e \in N$.

So, indeed the multiplication does "work out", and in fact, we can see that:

$gN = Ng$ for any $g \in G$ (this is obvious for $g \in N$, perhaps you can see why it works out for $g \in bN$). Indeed, arguing from the fact that "mod $N$" is an equivalence relation, we can see that $G$ has just two cosets, and since the left and right cosets of $eN = N = Ne$ coincide, the "only other coset" must coincide (left and right) for $g \not\in N$.

By contrast, let's look at the left cosets of $H$:

$H = \{e,b\}$
$aH = \{a,ab\}$
$a^2H = \{a^2,a^2b\}$.

If we are to have a group, we must have (since our group is of order $3$, the two non-identity elements must be inverses):

$(aH)(a^2H) = H$.

Again, we look at all $4$ possible products:

$a(a^2) = a^3 = e \in H$
$a(a^2b) = a^3b = b \in H$. So far, so good.
$(ab)(a^2) = a(ba)a = a(a^2b)a = a^3(ba) = ba = a^2b$. Huh. This ISN'T in $H$.
$(ab)(a^2b) = a(ba)ab = a(a^2b)ab = a^3(bab) = bab = a^2b^2 = a^2$. Neither is this.

The problem here is that $aH = \{a,ab\}$, but $Ha = \{a,ba\} = \{a,a^2b\} \neq aH$.

So we can't make the set of left cosets of $H$ into a group, it just doesn't work. The equivalence doesn't respect the group operation.

Now all of this is "at the element level", we're looking at the "nitty gritty" details of individual products. Let's look at the situation "from the outside".

With $N$, which happens to be normal, we can define in a perfectly acceptable way:

$(xN)(yN) = (xy)N$.

The above equation means that $x \mapsto [x]_{\sim_N} = xN$ is a homomorphism.

What does this homomorphism DO? For our $G$ and $N$, it basically says "let's turn all the $a$'s into the identity". We can see that this approach DOESN'T work for the $b$'s because it MATTERS which side (front or back) an $a$ is on with respect to $b$:

$ba = a^2b$ and $ab = ba^2$

So "turning $b$'s to $e$" would give $a = a^2$, which is not how things ARE in $G$.

But "turning $a$'s to $e$" DOES work, because it leads to $b = b$ (or $e = e$, if we just have $a$'s), which is fine.

So let's look at this situation just from knowing we have a congruence (normal subgroup induced equivalence), but don't know "the element details". We get a homomorphism:

$G \to G/N$ in which $G$ gets "condensed" ($N$ is "factored" out).

In this light, $\Bbb Z/5\Bbb Z$ is "identifying" $5$ with $0$, so we only get $5$ "virtually different" elements:

0 is 5 is 10 is...
1 is 6 is 11 is..
2 is 7 is 12...

and so on. All the "other" properties of the integers (commutativity, being cyclic) get "kept" we just have "five-ness" and 4 other kinds of "non-five-ness".

EVERY QUOTIENT GROUP ARISES IN THIS WAY, from an equivalence induced by a homomorphism. The equivalence is:

$g \sim g'$ if $\phi(g) = \phi(g')$ for our homomorphism $\phi:G \to \phi(G)$.

The normal subgroup we "quotient by" is $\text{ker }\phi = \phi^{-1}(e)$. It is straight-forward to show this is indeed a normal subgroup of $G$, and the existence of the (canonical) homomorphism $G \to G/N$ for any normal subgroup $N$ shows any normal subgroup gives rise to the kernel of a homomorphism.

I'll say this again, for emphasis: normal subgroups are kernels of homomorphisms.

Now, we are no longer talking about "elements", but just special kinds of mappings between groups. What kind of mappings? Surjective homomorphisms. More in post #2.

Thanks Deveno ... I really appreciate your help ...

Just working through your post now ... then with a better understanding of quotient groups, will go back to the Correspondence Theorem for Groups ... and when I have mastered that I will go back to the Correspondence Theorem for Rings ... where I was before I had to detour a bit ... ...:)

Thanks again for all your tutoring and help ...

* Just by the way, this post and the last would make great tutorials in the Math Resources section of the MHB site ... they are really helpful ... maybe others have trouble getting their head around quotient groups ...

Peter

 

[Deveno]

Now, let's look at $5\Bbb Z/10\Bbb Z$.

$5\Bbb Z = \{\dots, -20,-15,-10,-5,0,5,10,15,20\dots\}$ and

$10\Bbb Z = \{\dots,-30,-20,-10,0,10,20,30,\dots\}$.

Given $a + 10\Bbb Z$, we see it is:

$a + 10\Bbb Z = \{\dots,-30+a,-20+a,-10+a,a,10+a,20+a,30+a,\dots\}$.

Clearly, if $a$ is a multiple of $10$, then $a + 10\Bbb Z = 10\Bbb Z$, which is one of our possible cosets (left or right doesn't matter-the integers are an abelian group).

Well, that's HALF of the elements of $5\Bbb Z$, the other half are ODD multiples of $5$. So we have just two cosets:

$10\Bbb Z$ and $5 + 10\Bbb Z$, which means our quotient group is cyclic, of order $2$ (as all groups of order $2$ are).

We can look at this a different way:

The map $k \mapsto 5k$ is an isomorphism of $\Bbb Z$ with $5\Bbb Z$. Let's call it $\phi$.

Since $\phi$ is an isomorphism, it has an inverse isomorphism $\phi^{-1}$.

Note that we thus have:

$5\Bbb Z/10\Bbb Z \cong \phi^{-1}(5\Bbb Z)/\phi^{-1}(10\Bbb Z) = \Bbb Z/2\Bbb Z = \Bbb Z_2$

(see how much easier homomorphism make things?).

Never did get around to "quotients of quotients". Maybe tomorrow. :P

 

[Math Amateur]

Now, let's look at $5\Bbb Z/10\Bbb Z$.

$5\Bbb Z = \{\dots, -20,-15,-10,-5,0,5,10,15,20\dots\}$ and

$10\Bbb Z = \{\dots,-30,-20,-10,0,10,20,30,\dots\}$.

Given $a + 10\Bbb Z$, we see it is:

$a + 10\Bbb Z = \{\dots,-30+a,-20+a,-10+a,a,10+a,20+a,30+a,\dots\}$.

Clearly, if $a$ is a multiple of $10$, then $a + 10\Bbb Z = 10\Bbb Z$, which is one of our possible cosets (left or right doesn't matter-the integers are an abelian group).

Well, that's HALF of the elements of $5\Bbb Z$, the other half are ODD multiples of $5$. So we have just two cosets:

$10\Bbb Z$ and $5 + 10\Bbb Z$, which means our quotient group is cyclic, of order $2$ (as all groups of order $2$ are).

We can look at this a different way:

The map $k \mapsto 5k$ is an isomorphism of $\Bbb Z$ with $5\Bbb Z$. Let's call it $\phi$.

Since $\phi$ is an isomorphism, it has an inverse isomorphism $\phi^{-1}$.

Note that we thus have:

$5\Bbb Z/10\Bbb Z \cong \phi^{-1}(5\Bbb Z)/\phi^{-1}(10\Bbb Z) = \Bbb Z/2\Bbb Z = \Bbb Z_2$

(see how much easier homomorphism make things?).

Never did get around to "quotients of quotients". Maybe tomorrow. :P

Hi Deveno,

Just a small point of clarification:

Can you explain why/how it is the case that $$5\Bbb Z/10\Bbb Z \cong \phi^{-1}(5\Bbb Z)/\phi^{-1}(10\Bbb Z) $$?

Peter

 

[Deveno]

The inverse of an isomorphism is also an isomorphism.

Here's the underlying general fact:

Let $\phi: G \to G'$ be an isomorphism, and $K \lhd G$.

Then $G/K \cong \phi(G)/\phi(K)$.

Note that since $K \lhd G$, it follows that for any $k \in K$ and $g \in G$, that $gkg^{-1} \in K$. Since $\phi$ is a homomorphism, we have $\phi(g)\phi(k)\phi(g)^{-1} = \phi(gkg^{-1}) \in \phi(K)$. Since $\phi$ is bijective, every element of $\phi(K)$ can be written as $\phi(k)$ for a unique $k \in K$. Similarly, any element of $\phi(G)$ can be written as some $\phi(g)$, for some unique $g \in G$.Thus $\phi(K) \lhd \phi(G)$, so the group $\phi(G)/\phi(K)$ "makes sense".

I will show this by exhibiting an isomorphism $\psi$ between them.

Define $\psi(gK) = \phi(g)\phi(K)$.

As is usually the case when we use representatives (the "$g$") to define a mapping on cosets, we need to show $\psi$ is well-defined:

$gK = hK \implies \psi(gK) = \psi(hK)$.

Now $gK = hK \iff gh^{-1} \in K$. Therefore, $\phi(gh^{-1}) = \phi(g)\phi(h)^{-1} \in \phi(K)$.

Hence $\phi(g)\phi(K) = \phi(h)\phi(K)$ that is, $\psi(gK) = \psi(hK)$, and we have proven this assertion.

$\psi$ is injective: suppose $\psi(gK) = e_{\phi(G)/\phi(K)} = \phi(K)$.

This means $\phi(g)\phi(K) = \phi(K)$, i.e., $\phi(g) \in \phi(K)$. But $\phi(g)$ is the image of a unique $g \in G$, which must then lie in $K$.

Thus $gK = K$, the kernel of $\psi$ consists solely of the identity of $G/K$.

$\psi$ is surjective: let $x \in G'$ with $x\phi(K)$ any coset of $\phi(K)$. Since $\phi$ is an isomorphism, we have $x = \phi(a)$ for some unique $a \in G$. For this $a$, we have:

$\psi(aK) = \phi(a)\phi(K) = x\phi(K)$, which shows $\psi$ is surjective.

*************************

On a more informal level, isomorphisms preserve all the "abstract group" properties of a group, including the number and type of subgroups, the orders of elements, whether or not the group is abelian, the order of the group. It's much like translating a sentence from English to Spanish, faithfully.

 

[Math Amateur]

The inverse of an isomorphism is also an isomorphism.

Here's the underlying general fact:

Let $\phi: G \to G'$ be an isomorphism, and $K \lhd G$.

Then $G/K \cong \phi(G)/\phi(K)$.

Note that since $K \lhd G$, it follows that for any $k \in K$ and $g \in G$, that $gkg^{-1} \in K$. Since $\phi$ is a homomorphism, we have $\phi(g)\phi(k)\phi(g)^{-1} = \phi(gkg^{-1}) \in \phi(K)$. Since $\phi$ is bijective, every element of $\phi(K)$ can be written as $\phi(k)$ for a unique $k \in K$. Similarly, any element of $\phi(G)$ can be written as some $\phi(g)$, for some unique $g \in G$.Thus $\phi(K) \lhd \phi(G)$, so the group $\phi(G)/\phi(K)$ "makes sense".

I will show this by exhibiting an isomorphism $\psi$ between them.

Define $\psi(gK) = \phi(g)\phi(K)$.

As is usually the case when we use representatives (the "$g$") to define a mapping on cosets, we need to show $\psi$ is well-defined:

$gK = hK \implies \psi(gK) = \psi(hK)$.

Now $gK = hK \iff gh^{-1} \in K$. Therefore, $\phi(gh^{-1}) = \phi(g)\phi(h)^{-1} \in \phi(K)$.

Hence $\phi(g)\phi(K) = \phi(h)\phi(K)$ that is, $\psi(gK) = \psi(hK)$, and we have proven this assertion.

$\psi$ is injective: suppose $\psi(gK) = e_{\phi(G)/\phi(K)} = \phi(K)$.

This means $\phi(g)\phi(K) = \phi(K)$, i.e., $\phi(g) \in \phi(K)$. But $\phi(g)$ is the image of a unique $g \in G$, which must then lie in $K$.

Thus $gK = K$, the kernel of $\psi$ consists solely of the identity of $G/K$.

$\psi$ is surjective: let $x \in G'$ with $x\phi(K)$ any coset of $\phi(K)$. Since $\phi$ is an isomorphism, we have $x = \phi(a)$ for some unique $a \in G$. For this $a$, we have:

$\psi(aK) = \phi(a)\phi(K) = x\phi(K)$, which shows $\psi$ is surjective.

*************************

On a more informal level, isomorphisms preserve all the "abstract group" properties of a group, including the number and type of subgroups, the orders of elements, whether or not the group is abelian, the order of the group. It's much like translating a sentence from English to Spanish, faithfully.

Thanks Deveno ... but I need a little more help/clarification regarding the above post ...

You mention that we are trying to show that $G/K \cong \phi(G)/\phi(K)$ ... ...

... and then mention that you are going to exhibit an isomorphism between $G/K$ and $\phi(G)/\phi(K)$ ...

You then claim that $$\psi $$ is an isomorphism between $G/K$ and $\phi(G)/\phi(K)$ where $$\psi$$ is defined by:

$\psi(gK) = \phi(g)\phi(K)$

... ... BUT ...

... for $$\psi$$ to be an isomorphism between $G/K$ and $\phi(G)/\phi(K)$ it must map an element of $G/K$ onto an element of $\phi(G)/\phi(K)$ ...

BUT ... I cannot see why $\phi(g)\phi(K)$ is an element of $\phi(G)/\phi(K)$ ... ... $$\phi(g)$$ is certainly an element of $$\phi(G)$$ ... but how does $$\phi(K)$$ fit in ... I was expecting something like $$\phi(g)/\phi(k)$$ ... ...Can you clarify this matter and explain why it is the case that $\phi(g)\phi(K)$ is an element of $\phi(G)/\phi(K)$ ...
A further problem is that we need to show that $$\psi$$ is a homomorphism that is injective and surjective ... but we only seem to have shown the injectivity and the surjectivity ... can you clarify this matter also ...Hope you can help me further ...

Thanks for your extensive help so far ...

Peter

 

[Deveno]

The elements of the group $G/H$ are the cosets:

$H,g_1H,g_2H,g_3H,\dots$

for some set (such a complete set of representatives is called a "traversal" of $G$):

$\{g_i \in G\}$.

For example, $\{0,1,2,3,4\}$ is a traversal of $\Bbb Z/5\Bbb Z$.

Now the image of any group under a homomorphism is ALSO a group: the homomorphism property guarantees closure, and it is easy to verify that it also guarantees inverses, since:

$\phi(x)\phi(x^{-1}) = \phi(xx^{-1}) = \phi(e) = e_{\phi(G)}$

so that $\phi(x)^{-1} = \phi(x^{-1})$.

Now a subgroup $K$ of a group $G$ is also a group, so its image under $\phi$ is likewise a group, and since $\phi$ is ONTO $\phi(G)$ (it's the ENTIRE image) as a FUNCTION, the set $\phi(K)$ is a SUBSET of $\phi(G)$ that is a group in its own right. A subgroup.

Suppose $K = \{k_1,k_2,\dots,k_n\}$, for example.

Then $\phi(K) = \{\phi(k_1),\phi(k_2),\dots,\phi(k_n)\}$ (if $\phi$ is not injective, some of these will coincide).

Now for a fixed $g \in G$, we have:

$gK = \{gk_1,gk_2,\dots,gk_n\}$. So taking images under $\phi$ we have:

$\{\phi(gk_1),\phi(gk_2),\dots,\phi(gk_n)\}$ and by the homomorphism property we have:

$=\{\phi(g)\phi(k_1),\phi(g)\phi(k_2),\dots,\phi(g)\phi(k_n)\}$

$=\phi(g)\phi(K)$.

Here are two isomorphic groups:

$\Bbb Z_4$, under addition modulo $4$ and $\{1,i,-1,-i\}$ under complex multiplication.

Find the equivalent of $\{1,i,-1,-i\}/\langle -1\rangle$ for $\Bbb Z_4$.

Draw analogies between "evenness and oddness" and "real and imaginary" based on your investigation.

 

[Math Amateur]

The elements of the group $G/H$ are the cosets:

$H,g_1H,g_2H,g_3H,\dots$

for some set (such a complete set of representatives is called a "traversal" of $G$):

$\{g_i \in G\}$.

For example, $\{0,1,2,3,4\}$ is a traversal of $\Bbb Z/5\Bbb Z$.

Now the image of any group under a homomorphism is ALSO a group: the homomorphism property guarantees closure, and it is easy to verify that it also guarantees inverses, since:

$\phi(x)\phi(x^{-1}) = \phi(xx^{-1}) = \phi(e) = e_{\phi(G)}$

so that $\phi(x)^{-1} = \phi(x^{-1})$.

Now a subgroup $K$ of a group $G$ is also a group, so its image under $\phi$ is likewise a group, and since $\phi$ is ONTO $\phi(G)$ (it's the ENTIRE image) as a FUNCTION, the set $\phi(K)$ is a SUBSET of $\phi(G)$ that is a group in its own right. A subgroup.

Suppose $K = \{k_1,k_2,\dots,k_n\}$, for example.

Then $\phi(K) = \{\phi(k_1),\phi(k_2),\dots,\phi(k_n)\}$ (if $\phi$ is not injective, some of these will coincide).

Now for a fixed $g \in G$, we have:

$gK = \{gk_1,gk_2,\dots,gk_n\}$. So taking images under $\phi$ we have:

$\{\phi(gk_1),\phi(gk_2),\dots,\phi(gk_n)\}$ and by the homomorphism property we have:

$=\{\phi(g)\phi(k_1),\phi(g)\phi(k_2),\dots,\phi(g)\phi(k_n)\}$

$=\phi(g)\phi(K)$.

Here are two isomorphic groups:

$\Bbb Z_4$, under addition modulo $4$ and $\{1,i,-1,-i\}$ under complex multiplication.

Find the equivalent of $\{1,i,-1,-i\}/\langle -1\rangle$ for $\Bbb Z_4$.

Draw analogies between "evenness and oddness" and "real and imaginary" based on your investigation.

Hi Deveno,

Thanks for the help ...

You write:

" ... ... Here are two isomorphic groups:

$\Bbb Z_4$, under addition modulo $4$ and $\{1,i,-1,-i\}$ under complex multiplication.

Find the equivalent of $\{1,i,-1,-i\}/\langle -1\rangle$ for $\Bbb Z_4$. ... ... "I am struggling with this problem ... can you help with determining the function that is the isomorphism ... ?Peter

- - - Updated - - -

The elements of the group $G/H$ are the cosets:

$H,g_1H,g_2H,g_3H,\dots$

for some set (such a complete set of representatives is called a "traversal" of $G$):

$\{g_i \in G\}$.

For example, $\{0,1,2,3,4\}$ is a traversal of $\Bbb Z/5\Bbb Z$.

Now the image of any group under a homomorphism is ALSO a group: the homomorphism property guarantees closure, and it is easy to verify that it also guarantees inverses, since:

$\phi(x)\phi(x^{-1}) = \phi(xx^{-1}) = \phi(e) = e_{\phi(G)}$

so that $\phi(x)^{-1} = \phi(x^{-1})$.

Now a subgroup $K$ of a group $G$ is also a group, so its image under $\phi$ is likewise a group, and since $\phi$ is ONTO $\phi(G)$ (it's the ENTIRE image) as a FUNCTION, the set $\phi(K)$ is a SUBSET of $\phi(G)$ that is a group in its own right. A subgroup.

Suppose $K = \{k_1,k_2,\dots,k_n\}$, for example.

Then $\phi(K) = \{\phi(k_1),\phi(k_2),\dots,\phi(k_n)\}$ (if $\phi$ is not injective, some of these will coincide).

Now for a fixed $g \in G$, we have:

$gK = \{gk_1,gk_2,\dots,gk_n\}$. So taking images under $\phi$ we have:

$\{\phi(gk_1),\phi(gk_2),\dots,\phi(gk_n)\}$ and by the homomorphism property we have:

$=\{\phi(g)\phi(k_1),\phi(g)\phi(k_2),\dots,\phi(g)\phi(k_n)\}$

$=\phi(g)\phi(K)$.

Here are two isomorphic groups:

$\Bbb Z_4$, under addition modulo $4$ and $\{1,i,-1,-i\}$ under complex multiplication.

Find the equivalent of $\{1,i,-1,-i\}/\langle -1\rangle$ for $\Bbb Z_4$.

Draw analogies between "evenness and oddness" and "real and imaginary" based on your investigation.

Hi Deveno,

Thanks again for your extensive help ...

You write:

" ... ... $$\{\phi(g)\phi(k_1),\phi(g)\phi(k_2),\dots,\phi(g)\phi(k_n)\}

=\phi(g)\phi(K)$$ ... "

I assume that this statement is essentially the definition of $\phi(g)\phi(K)$ ... ... that is it is "true" by nature of being a definition ... can you confirm ...

Peter