Determining Quotient Group (\mathbb{Z}_2\times\mathbb{Z}_4)/\langle(1,2)\rangle

[Oxymoron]

Question 1
Determine the quotient group $$(\mathbb{Z}_2\times\mathbb{Z}_4)/\langle(1,2)\rangle$$

Answer
$$\langle(1,2)\rangle$$ is a cyclic subgroup $$H$$ of $$\mathbb{Z}_2\times\mathbb{Z}_4$$ generated by $$(1,2)$$. Thus

$$H=\{(0,0),(1,2)\}$$

Since $$\mathbb{Z}_2\times\mathbb{Z}_4$$ has 2.4 = 8 elements, and $$H$$ has 2 elements, all cosets of $$H$$ must have 2 elements, and $$(\mathbb{Z}_2\times\mathbb{Z}_4)/H$$ must have order 4.

Possible abelian groups of order 4 are

$$\mathbb{Z}_2\times\mathbb{Z}_2$$
$$\mathbb{Z}\times\mathbb{Z}_4$$

But I don't know how to work out which one is isomorphic to $$(\mathbb{Z}_2\times\mathbb{Z}_4)/\langle(1,2)\rangle$$

 

[Data]

Which of $$\mathbb{Z}_2 \times \mathbb{Z}_2$$, $$\mathbb{Z}_1 \times \mathbb{Z}_4$$, and $$(\mathbb{Z}_2 \times \mathbb{Z}_4)/H$$ are cyclic?

 

[Oxymoron]

I think $$\mathbb{Z}\times\mathbb{Z}_4$$ is cyclic because it is generated by $$a^n$$ for all $$n=0,1,\dots$$ where $$a=(1,2)$$ and $$n = 2$$

 

[Data]

Well, as I posted, you really mean $\mathbb{Z}_1 \times \mathbb{Z}_4$ (or equivalently just $\mathbb{Z}_4$), since $\mathbb{Z} \times \mathbb{Z}_4$ doesn't have order 4 at all (not even finite order!).

But it is indeed cyclic (generated by $(0, 1)$ or just $1$ if you choose the $\mathbb{Z}_4$ variety~).

Is $(\mathbb{Z}_2 \times \mathbb{Z}_4)/H$ cyclic?

 

[Oxymoron]

Yes, it is cyclic. Isn't there a theorem that says a quotient group of cyclic group is cyclic?

So wouldn't $$\mathbb{Z}\times\mathbb{Z}_4$$ be the answer? Or am I missing something?

 

[Data]

I don't remember such a theorem. Let's just write down the elements of [itex](\mathbb{Z}_2 \times \mathbb{Z}_4)/H[/tex]. This set is just

$$\{ (1, 0) + H, \; (1, 1) + H, \; (0, 1) + H, \; (0, 0) + H \}$$.

Does it have a generator? Remember the group is abelian and $H + H = H$.

Edit: Actually, there may be. I really don't remember! Anyways, in this case the group in question is indeed cyclic (and thus isomorphic to $\mathbb{Z}_4$).

 

[Oxymoron]

$$\mathbb{Z}_2\times\mathbb{Z}_4 = \{(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)\}$$

But this can be collapsed into

$$\mathbb{Z}_4 = \{(0,0),(0,1),(0,2),(0,3)\} = \mathbb{Z}_2\times\mathbb{Z}_2$$

So then

$$(\mathbb{Z}_2\times\mathbb{Z}_4)/H = \{(0,0)+H,(0,1)+H,(0,2)+H,(0,3)+H\}$$

where $$H=\langle(1,2)\rangle$$

and its generator is $$(1,2)$$

 

[Data]

I think the theorem you're thinking of is that if $G$ is cyclic then every quotient group $G/H$ is cyclic. In our case we can't use that (since $H$ is cyclic, not $\mathbb{Z}_2 \times \mathbb{Z}_4$). There's no theorem the way we need it

 

[Data]

I wouldn't include the line saying "can be collapsed into" (what does that mean?). And I think in your second TeX line (the one I think you should remove, because the first equality isn't true~) you meant $= \mathbb{Z}_1 \times \mathbb{Z}_4$, not $= \mathbb{Z}_2 \times \mathbb{Z}_2$!

The rest looks good though~

 

[Oxymoron]

Ok, I fixed it up a bit.

Let me get this straight. The quotient group under investigation is
$$(\mathbb{Z}_2\times\mathbb{Z}_4)/\langle(1,2)\rangle$$.

Here $$\langle(1,2)\rangle$$ is the cyclic subgroup $$H$$ of $$\mathbb{Z}_2\times\mathbb{Z}_4$$ generated by $$(1,2)$$. Thus

$$H = \{(0,0),(1,2)\}$$

Since $$\mathbb{Z}_2\times\math{Z}_4$$ has 8 elements and $$H$$ has 2 elements, all cosets of $$H$$ must have 4 elements, and $$(\mathbb{Z}_2\times\mathbb{Z}_4)/H$$.

In additive notation, the cosets are

$$H = \{(0,0)+H,(0,1)+H,(0,2)+H,(0,3)+H\}$$

Since we can compute by choosing representatives $$(0,0),(0,1),(0,2),(0,3)$$ it is clear that $$(\mathbb{Z}_2\times\mathbb{Z}_4)/H$$ is isomorphic to $$\mathbb{Z}_4$$

Note that this is what we would expect, since in a factor modulo $$H$$, everything in $$H$$ becomes the identity element; that is we are essentially setting everything in $$H$$ to zero. Thus the whole factor $$\mathbb{Z}_2$$ of $$\mathbb{Z}_2\times\mathbb{Z}_4$$ is collapsed, leaving just the second factor $$\mathbb{Z}_4$$.

 

[Data]

looks good

 

[Oxymoron]

Question 2

Show that ifa group $$G$$ has exactly one subgroup $$H$$ of given order, then this subgroup must be normal.

I figured that since every group has at least two subgroups: the whole group itself (the improper subgroup) $$G$$ and the trivial subgroup $$\{e\}$$.

If a group $$G$$ has exactly one subgroup then $$G = \{e\}$$ and the group in question is actually the trivial group. This is the only way that I can see that a group can have exactly one subgroup - if it is the trivial group.

So if $$G$$ is the trivial group, then it consists of only one element, namely the identity. The group consisting of one element, $$e$$ is commutative. ie $$e*a=a*e$$ where $$a \in G$$. Obviously $$a=e$$ if $$G$$ has only one element.

Hence $$G$$ is abelian. And all subgroups of abelian groups are normal.

Proof

A subgroup is normal if its left and right cosets coincide. That is, $$gH=Hg \, \forall \, g \in G$$. Since $$g=e$$ then we have $$eH=He$$ which is trivial because $$G$$ is abelian.

Therefore the subgroup $$H$$ of $$G$$ is normal.

 

[Data]

You are misinterpreting the question (at least the way I read it). It is not saying that $G$ has only one subgroup. It says that $G$ has only one subgroup $H$ of a particular order, say $n$, ie. if $J \leq G$ and $|J| = n$ then $J = H$.

 

[Oxymoron]

Ok, let's assume that you read the question correctly (which is likely). Then I have to show that the subgroup is normal given that it is the only subgroup with a particular order $$n$$.

This means that $$G$$ has only one subgroup with $$n$$ elements, all other subgroups have some other order.

For a subgroup of a group to be normal, then the following are equivalent
$$ghg^{-1} \in H \, \forall \, g\in G \mbox{and} h\in H$$
$$gHg^{-1} = H \, \forall \, g\in G$$
$$gH = Hg \, \forall g\in G$$

But I don't know how to start. That is, none of the conditions for a subgroup to be normal, involve the order of the subgroup.

Any hints?

 

[fourier jr]

also
N normal <==> NaNb = Nab for a, b not in subgroup N

 

[Data]

Sorry I haven't replied in a while.

Here's a hint:

Recall that $H$ normal in $G$ if and only if $gHg^{-1} = H \mbox{ for every } g \ \mbox{in} \ G$. Assume $H$ isn't normal. Then there is some $g$ in $G$ such that $gHg^{-1} \neq H$. For this $g$, consider the set $gHg^{-1}$ (Hint: Try to prove that it is a subgroup of $G$ of the same order as $H$, to get a contradiction).

 

[Data]

Figured I'd put this back at the top in case you still need it, since you're online right now~

 

[Oxymoron]

The condition for $$H$$ being normal is

$$ghg^{-1} \in H\quad \forall g \in G,\, \forall h \in H$$

Assume that $$H$$ isn't normal (proof by contradiction)

Then $$\exists \, k \in G$$ such that

$$khk^{-1} \notin H \quad \forall h \in H$$

Now consider the set

$$A = \{khk^{-1} | \, \forall h \in H\}$$

We now show that this is a subgroup of $$G$$ and that it has the same order as $$H$$.

For $$h_1,h_2 \in H, \, \exists\, kh_1k^{-1},\, kh_2k^{-1} \in A$$ by definition. Now,

$$(kh_1k^{-1})(kh_2k^{-1}) = (kh_1)(k^{-1}k)(h_2k^{-1})$$
$$= kh_1(e)h_2k^{-1}$$
$$= k(h_1h_2)k^{-1}$$

But $$h_1,h_2 \in H$$ as $$H$$ is a subgroup by assumption.

Hence $$kh_1,h_2k^{-1} \in A$$ which implies that $$A$$ is a subgroup of $$G$$

Now so far we have been using the fact that $$H$$ is not normal. And from this we have got a subgroup $$A$$. But it is obvious that for each $$h \in H$$ we can make an $$a \in A$$ by

$$kh_1k^{-1} = kh_2k^{-1} \Leftrightarrow h_1 = h_2$$

the left and right cancellation laws. Hence $$H$$ and $$A$$ have the same order. Alternatively I think I could have proven this by setting a bijection. Anyway, $$H$$ and $$A$$ having the same order is a contradiction since $$H$$ is the only subgroup with a given order. Hence $$H$$ MUST be normal.

 

[Data]

Everything looks great, except for one detail:

You cannot assume $\exists k \in G$ such that

$$khk^{-1} \notin H \; \forall h \in H$$.

This is too strong.

All you can assume is

$\exists k \in G$ such that

$$\{ khk^{-1} | h \in H \} \neq \{ h \in H\}$$

there is a subtle difference, but it does not affect the rest of the proof.

Also, you must show that $e \in A$, but this is trivial. Also, that $x \in A \Longrightarrow x^{-1} \in A$ (also easy).

You are quite correct that the trivial bijection (and, in fact, isomorphism)

$$f : H \longrightarrow A$$
$$\forall \ h \in H, \ f(h) = khk^{-1}$$

would also have worked nicely

 

[Oxymoron]

Thanks Data for your input. I'd have to say, you have a knack for mentoring.

Here is another question I have been working on.

Let $$\textsc{Q}$$ be the subgroup of $$GL(2,\mathbb{C})$$ generated by

$$\left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array}\right) \quad \mbox{and} \quad \left(\begin{array}{cc} 0 & i \\ i & 0 \\ \end{array}\right)$$

1. Show that $$\textsc{Q}$$ is a nonabelian group of order 8.
2. Is $$\textsc{Q}$$ isomorphic to the dihedral group $$\mathcal{D}_4$$?

 

[Oxymoron]

For 1. I let a equal the first matrix and b equal the second. Then I used Maple to calculate all possible combinations.

a, aa, aaa, aaaa, b, bb, bbb, bbbb, ab, aab, aaab, abb, abbb, etc...

I eventually discovered that there is only 8 different answers.

Is there any quicker way of determining the order to be 8? Because as it stands I have pages of matrix calculations, showing that only 8 matrices exist after all possible combinations of multiplying the two together. Very time consuming.

As well as this I discovered that $$ab \neq ba$$

Therefore the subgroup is not commutative for all a and b, and hence is not abelian.

 

[Data]

This group actually has a special name, specifically, the quaternions. In algebraic form they are very interesting (they are an extension to the complex numbers, in which $-1$ has 3 square roots!). They make cross and dot products of 3-vectors very natural to take, for example. You can read about them here: http://mathworld.wolfram.com/Quaternion.html

I'm curious, what textbook are you using for this class? I think those questions are verbatim from the one I used in my first abstract algebra class

Anyways, on to your specific question. There is indeed a better way to verify that the order is 8. Let


$$A = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right)$$

$$B = \left(\begin{array}{cc} 0 & i \\ i & 0\end{array}\right)$$

Just calculate $A^2$, $B^2$, and $BA$, and it should be obvious how to proceed (you should find $A^2 = B^2 = -I$ where $I$ is the 2x2 identity matrix, and $BA = -AB$).

The reason to do this is that you will no longer have to do any actual matrix calculations. These three facts are the key to every multiplication you can do (eg. from there I can immediately say $AB^3 = -B^3A = -(-B)A=BA$).

A nice form for the set you end up with is

$$\{I, A, A^2, A^3, B, BA, BA^2, BA^3\}$$

 

[Oxymoron]

$$Q$$ is isomorphic to $$\mathcal{D}_4$$.

Let $$\star$$ denote the matrix multiplication binary operation, and $$\star'$$ be the binary operation on dihedral groups (rotation and reflection).

Let $$A, B \in Q$$ and $$a,b \in \mathcal{D}_4$$. Then if $$\phi : Q \rightarrow H$$ is an isomorphism...

$$A^4 \star B^2 = \phi(a^4) \star' \phi(b^2)$$

...must hold.

Since $$A^4 = B^2 = e$$ and by the definition of Dihedral groups we know $$a^4 = b^2 = e$$, hence LHS = RHS = e. And thus $$Q$$ is isomorphic to $$\mathcal{D}_4$$.

 

[Data]

They actually aren't isomorphic. They look like it, but they aren't quite (I actually made the same mistake at one point ). Note that in ${\cal{D}} _4$, and letting $\rho$ be reflection and $R$ be rotation, we have three elements of order 2: $R^2, \ \rho,$ and $\rho R^2$.

In $Q$ we have only one: $A^2$.

 

[Oxymoron]

So since D4 has three elements of order 2 and Q has only one, the two can obviously not be bijective? So my proof was correct but not strong enough to encompass all properties?

 

[Data]

Well, there's definitely a bijection between the two (they have the same order, which by definition means there's a bijection between them). There's no isomorphism (recall that an isomorphism is a bijection that is also a homomorphism). Everything in your previous post was right, up until the last sentence.

 

[Muzza]

$$Q$$ is isomorphic to $$\mathcal{D}_4$$.

Let $$\star$$ denote the matrix multiplication binary operation, and $$\star'$$ be the binary operation on dihedral groups (rotation and reflection).

Let $$A, B \in Q$$ and $$a,b \in \mathcal{D}_4$$. Then if $$\phi : Q \rightarrow H$$ is an isomorphism...

$$A^4 \star B^2 = \phi(a^4) \star' \phi(b^2)$$

...must hold.

Since $$A^4 = B^2 = e$$ and by the definition of Dihedral groups we know $$a^4 = b^2 = e$$, hence LHS = RHS = e. And thus $$Q$$ is isomorphic to $$\mathcal{D}_4$$.

This confuses me greatly. You assume there is an isomorphism between Q and D_4, and using that, you conclude that that they are isomorphic. Doesn't this seem circular?

 

[Oxymoron]

Muzza, not quite. Here was my line of thinking.

Suppose isomorphism then __ must hold. Prove __ holds then it must be an isomorphism.

 

[Oxymoron]

Well, there's definitely a bijection between the two (they have the same order, which by definition means there's a bijection between them). There's no isomorphism (recall that an isomorphism is a bijection that is also a homomorphism). Everything in your previous post was right, up until the last sentence.

Data, did my proof prove that $$\phi$$ is an isomorphism or a homomorphism? I'm just looking at the definition of a homomorphism and it says that

$$\phi(ab) = \phi(a)\phi(b)$$

If not, then how do I prove that $$\phi$$ is not an isomorphism (that Q and D4 are isomorphic). I'm having a hard time coming to terms with this since I just convinced myself that they ARE isomorphic.

 

[Muzza]

Suppose isomorphism then __ must hold. Prove __ holds then it must be an isomorphism.

You're saying that if P implies Q, then Q implies P. But that's not true in general. My car is white. Now, if I see a white car, can I conclude that the car is mine?

A mathematical example: if an infinite sum $\sum_{k = 1}^{\infty} a_k$ converges, then a_n tends to 0. But if a_n is a sequence that tends to 0, it's not necessarily true that $\sum_{k = 1}^{\infty} a_k$ converges (consider a_n = 1/n).

Data, did my proof prove that phi is an isomorphism or a homomorphism?

But you assumed phi was an isomorphism! One could easily prove any statement (Poincaré conjecture, Fermat's last theorem, you name it) if one was allowed to assume the truth of the statement in one's "proof"...

No, first you should prove that if G and H are groups and f: G -> H is an isomorphism, then the order of f(x) is equal to the order of x (where x is any element in G).

Then, suppose there was an isomorphism f between D_4 and Q. As Data pointed out, there are two elements x, y of order 2 in D_4. Then f(x) and f(y) would also be of order 2. Contradiction, since there is only one element of order 2 in Q.

 

[Data]

Just because

$$\phi(ab) = \phi(a)\phi(b)$$

for some particular $a$ and $b$ does not imply that it works for every $a$ and $b$. This is the problem. A homomorphism requires it to work for any choices. Thus you didn't prove that anything was a homomorphism or a bijection (although it is certainly possible to find bijections and homomorphisms between the two. It is not possible to find an isomorphism, ie. a function that is both a bijection and a homomorphism. This is what you need to prove).

And actually I spoke a little too soon in my last post (I need to read more carefully!). In $Q$, we find $A^4B^2 =e(-e) = - e \neq e$ but in ${\cal D}_4$ we find $R^4 \rho^2 = ee = e$. So your example doesn't quite work anyways.

The most clear path to prove what you need is that that Muzza suggested:

Prove that if $f(x)$ is an isomorphism between to groups then $\mbox{ord}f(x) = \mbox{ord}x$ and then use the fact that I gave above of unequal numbers of elements of certain orders to get your result.

 

[Oxymoron]

Thanks Data. That is what I wanted to hear. And Muzza's post was most helpful too. I will post back with another attempt later.

Thankyou guys.

By the way Muzza, your "framework" proof, is exactly the path I was trying to take (same idea). Except mine came out all fuzzy.

 

[Oxymoron]

If $$Q$$ and $$\mathcal{D}_4$$ are groups, and $$\phi : Q \rightarrow \mathcal{D}_4$$ is an isomorphism. Then for all $$x \in Q$$,

$$|\phi(x)| = |x|$$

That is, $$\phi$$ maps from exactly one element in $$Q$$ to exactly one element in $$\mathcal{D}_4$$.

Suppose that $$\phi$$ is an isomorphism. Denote $$a$$ to be rotation and $$b$$ to be reflection in $$\mathcal{D}_4$$. Then there are three elements in $$\mathcal{D}_4$$ with order 2:

$$a^2, b, ab^2$$

In $$Q$$ there is only one element of order 2:

$$A^2$$

where A is the 2x2 matrix that I typed earlier.

So if $$|A| = 2$$ then

$$|\phi(A)| = |a^2| = |b| = |ab^2| \in \mathcal{D}_4$$

Thus $$\phi$$ is not an isomorphism.

 

[Oxymoron]

I want to look closer into the dihedral group $$\mathcal{D}_n$$. Where

$$\mathcal{D}_n = \{a,b | a^n = b^2 = e, \, bab = a^{-1}\}$$.

I want to find all normal subgroups and then determine the corresponding quotient groups.


A subgroup $$H$$ of a group $$G$$ is normal if

$$gHg^{-1} = H \quad \forall \, g \in G$$

Take $$n$$ to be odd. If we apply rotation $$a$$ to the n-gon in the positive sense, and proceed to apply it in the negative sense, then we haven't rotated the n-gon at all.

Im talking jibberish. How am I supposed to find the normal subgroups of this group? What are they going to look like? Subsets? I don't know...

 

[Oxymoron]

Note, that if this was a specific dihedral group $$\mathcal{D}_3$$ or something. Then I would write up the multiplication table, work out the subgroups, and then determine which are normal by applying

$$gH = Hg \, \forall \, g\in G$$

But I have no idea how to start doing this problem when $$n$$ is arbitrary.