MHB Show that it is a normal subgroup of S4

[mathmari]

Hey!

I want to show that $N\{1, (12)(34), (13)(24), (14)(23)\}$ is a normal subgroup of $S_4$ that is contained in $A_4$ and that satisfies $S_4/N\cong S_3$ and $A_4/N\cong Z_3$. Let $\sigma\in S_4$.

We have the following:
$$\sigma 1 \sigma^{-1}=\sigma (1) \\ \sigma (1 2)(3 4) \sigma^{-1}=\left (\sigma (1) \sigma (2)\right ) \left (\sigma(3) \sigma (4)\right ) \\ \sigma (13)(24) \sigma^{-1}=\left (\sigma (1) \sigma (3)\right )\left (\sigma (2)\sigma (4)\right ) \\ \sigma (14)(23) \sigma^{-1}=\left (\sigma (1)\sigma (4)\right )\left (\sigma (2)\sigma (3)\right )$$

Right? (Wondering)

But how could we show that these are elements of $N$ ? (Wondering)

 

[johng1]

An "easy" counting problem: the number of elements of $S_4$ that are the product of 2 disjoint 2 cycles is 3. So $N$ contains all such elements. As you have shown, conjugation preserves the length and number of cycles in any element. So $N$ is normal in $S_4$.
2 cycles are odd permutations; so the product of 2 2 cycles is even. Hence every element of $N$ is an even permutation; i.e. $N\subseteq A_4$.
$A_4/N$ is of order 3 and hence isomorphic to $Z_3$
$S_4/N$ is of order 6. Since there are only two groups of order 6 (cyclic and $S_3$), you just have to show $S_4/N$ is not cyclic. Hint: find more than 1 element of order 2 in $S_4/N$.

 

[I like Serena]

Hey!

I want to show that $N\{1, (12)(34), (13)(24), (14)(23)\}$ is a normal subgroup of $S_4$ that is contained in $A_4$ and that satisfies $S_4/N\cong S_3$ and $A_4/N\cong Z_3$. Let $\sigma\in S_4$.

We have the following:
$$\sigma 1 \sigma^{-1}=\sigma (1) \\ \sigma (1 2)(3 4) \sigma^{-1}=\left (\sigma (1) \sigma (2)\right ) \left (\sigma(3) \sigma (4)\right ) \\ \sigma (13)(24) \sigma^{-1}=\left (\sigma (1) \sigma (3)\right )\left (\sigma (2)\sigma (4)\right ) \\ \sigma (14)(23) \sigma^{-1}=\left (\sigma (1)\sigma (4)\right )\left (\sigma (2)\sigma (3)\right )$$

Right? (Wondering)

But how could we show that these are elements of $N$ ? (Wondering)

Hey mathmari! (Smile)

I'd say that $\sigma 1 \sigma^{-1}=\sigma \sigma^{-1} = 1$, which is contained in $N$.

As for the others, I believe they are each 2 disjoint 2-cycles.
Are there any such permutations in $S_4$ that are missing in $N$? (Wondering)

 

[mathmari]

An "easy" counting problem: the number of elements of $S_4$ that are the product of 2 disjoint 2 cycles is 3. So $N$ contains all such elements. As you have shown, conjugation preserves the length and number of cycles in any element. So $N$ is normal in $S_4$.

I'd say that $\sigma 1 \sigma^{-1}=\sigma \sigma^{-1} = 1$, which is contained in $N$.

As for the others, I believe they are each 2 disjoint 2-cycles.
Are there any such permutations in $S_4$ that are missing in $N$? (Wondering)

All the possible 2-cycles of $S_4$ are the following:
$$\{\left (\sigma (1) \sigma (2)\right ) \left (\sigma(3) \sigma (4)\right ) ,\left (\sigma (1) \sigma (3)\right )\left (\sigma (2)\sigma (4)\right ) ,\left (\sigma (1)\sigma (4)\right )\left (\sigma (2)\sigma (3)\right )\} \tag 1$$

And since all the possible 2-cycles of (1 2 3 4) are the following:
$$\{(12)(34), (13)(24), (14)(23)\} \tag 2$$
so the elements of the set $(1)$ and the elements of the set $(2)$ must be the same.

Therefore, we have the following:
$$\sigma 1 \sigma^{-1}=\sigma \sigma^{-1}=1\in N \\ \sigma (1 2)(3 4) \sigma^{-1}=\left (\sigma (1) \sigma (2)\right ) \left (\sigma(3) \sigma (4)\right ) \in N\\ \sigma (13)(24) \sigma^{-1}=\left (\sigma (1) \sigma (3)\right )\left (\sigma (2)\sigma (4)\right ) \in N\\ \sigma (14)(23) \sigma^{-1}=\left (\sigma (1)\sigma (4)\right )\left (\sigma (2)\sigma (3)\right ) \in N$$

Is everything correct? (Wondering)

 

[I like Serena]

All the possible 2-cycles of $S_4$ are the following:
$$\{\left (\sigma (1) \sigma (2)\right ) \left (\sigma(3) \sigma (4)\right ) ,\left (\sigma (1) \sigma (3)\right )\left (\sigma (2)\sigma (4)\right ) ,\left (\sigma (1)\sigma (4)\right )\left (\sigma (2)\sigma (3)\right )\} \tag 1$$

Erm... those aren't 2-cycles are they? (Wondering)
They are permutations that are 2 disjoint 2-cycles.

And isn't it about them being part of $N$ rather than $S_4$? (Wondering)

 

[mathmari]

Erm... those aren't 2-cycles are they? (Wondering)
They are permutations that are 2 disjoint 2-cycles.

(Blush)

The 2-cycles are the following:
$$(1 2), (1 3), (1 4), (2 3), (2 4), (3 4)$$
or not? (Wondering)

 

[I like Serena]

(Blush)

The 2-cycles are the following:
$$(1 2), (1 3), (1 4), (2 3), (2 4), (3 4)$$
or not? (Wondering)

Yep! Those are all possible 2-cycles in $S_4$! (Nod)

 

[mathmari]

I got stuck right now...

Why are $$\sigma (1 2)(3 4) \sigma^{-1}=\left (\sigma (1) \sigma (2)\right ) \left (\sigma(3) \sigma (4)\right ) \\ \sigma (13)(24) \sigma^{-1}=\left (\sigma (1) \sigma (3)\right )\left (\sigma (2)\sigma (4)\right ) \\ \sigma (14)(23) \sigma^{-1}=\left (\sigma (1)\sigma (4)\right )\left (\sigma (2)\sigma (3)\right )$$ elements of $N$ ? (Wondering)

 

[I like Serena]

I got stuck right now...

Why are $$\sigma (1 2)(3 4) \sigma^{-1}=\left (\sigma (1) \sigma (2)\right ) \left (\sigma(3) \sigma (4)\right ) \\ \sigma (13)(24) \sigma^{-1}=\left (\sigma (1) \sigma (3)\right )\left (\sigma (2)\sigma (4)\right ) \\ \sigma (14)(23) \sigma^{-1}=\left (\sigma (1)\sigma (4)\right )\left (\sigma (2)\sigma (3)\right )$$ elements of $N$ ? (Wondering)

Each of them are 2 disjoint 2-cycles.
And $N$ contains all 2 disjoint 2-cycles that $S_4$ contains... (Thinking)

 

[mathmari]

Each of them are 2 disjoint 2-cycles.
And $N$ contains all 2 disjoint 2-cycles that $S_4$ contains... (Thinking)

Ah ok... I see... (Thinking)

2 cycles are odd permutations; so the product of 2 2 cycles is even. Hence every element of $N$ is an even permutation; i.e. $N\subseteq A_4$.

We have that $N\{1, (12)(34), (13)(24), (14)(23)\}$. Is $1$ also an even permutation? (Wondering)
All of $(12)(34), (13)(24), (14)(23)$ are even because we have at each case an even number of transpositions, $2$, right? (Wondering)

 

[I like Serena]

We have that $N\{1, (12)(34), (13)(24), (14)(23)\}$. Is $1$ also an even permutation? (Wondering)
All of $(12)(34), (13)(24), (14)(23)$ are even because we have at each case an even number of transpositions, $2$, right? (Wondering)

Yes, and indeed $1$ is even as well - it consists of $0$ transpositions. (Nerd)

 

[mathmari]

Yes, and indeed $1$ is even as well - it consists of $0$ transpositions. (Nerd)

Ah ok... (Thinking)

$A_4/N$ is of order 3 and hence isomorphic to $Z_3$
$S_4/N$ is of order 6. Since there are only two groups of order 6 (cyclic and $S_3$), you just have to show $S_4/N$ is not cyclic. Hint: find more than 1 element of order 2 in $S_4/N$.

How do we see that $A_4/N$ is of order 3 and $S_4/N$ is of order 6? (Wondering)

Could we show that $A_4/N$ is isomorphic to $Z_3$ by defining a function and showing that it is bijective and an homomorphism. And the same for $S_4/N\cong S_3$ ? (Wondering)

 

[johng1]

You know that the order of $S_4=24$ and the order of $A_4=12$. So just compute the order of a factor group as the order of the "numerator" divided by the order of the "denominator".For your second question, yes, but why work so hard?

 

[Deveno]

Another fundamental result on groups:

For any finite group $G$, the (left or right) cosets of a subgroup $H$ form a partition of $G$, so:

$G = \bigcup\limits_{g\in G} gH$.

If we only count each coset once, that is, we count only *distinct* cosets, then this is a disjoint union:

$G = \coprod\limits_i g_iH$

so that $|G| = |H| + |g_1H| + |g_2H| + \cdots + |g_{n-1}H|$, where each of these (left) cosets is distinct, and $G$ has $n$ cosets.

Since the mapping $L_{g_i}:G \to G$ given by $L_{g_i}(x) = g_ix$ is bijective for every $g_i \in G$, we have for any $g_i$ that $|g_iH| = |H|$ and so:

$|G| = n\cdot|H|$.

This is usually how Lagrange's theorem is proven. The number $n$ (the number of cosets) is called the *index* of $H$ in $G$, and is written $[G]$. We thus have:

$|G| = [G]\cdot |H|$.

If $H$ is a normal subgroup, so that $G/H$ is a group under coset multiplication, we have:

$|G/H| = [G] = \dfrac{|G|}{|H|}$.

 

[mathmari]

We have that $|S_4|=4!=24$ and $|N|=4$ (since $N$ contains $4$ elements), then $|S_4/N|=\frac{|S_4|}{|N|}=\frac{24}{4}=6$, since we have shown that $N$ is a normal subgroup of $S_4$.

The only groups of order $6$ are the cyclic one and $S_3$.

$S_4/N$ is of order 6. Since there are only two groups of order 6 (cyclic and $S_3$), you just have to show $S_4/N$ is not cyclic. Hint: find more than 1 element of order 2 in $S_4/N$.

Why do we have to find more than 1 element of order 2 in $S_4/N$ to show that $S_4/N$ is not cyclic?

If $H$ is a normal subgroup, so that $G/H$ is a group under coset multiplication, we have:

$|G/H| = [G] = \dfrac{|G|}{|H|}$.

To apply this formula at the calculation of the order of $A_4/N$ do we have to show that $N$ is a normal subgroup of $A_4$ ? (Wondering)

 

[Deveno]

We have that $|S_4|=4!=24$ and $|N|=4$ (since $N$ contains $4$ elements), then $|S_4/N|=\frac{|S_4|}{|N|}=\frac{24}{4}=6$, since we have shown that $N$ is a normal subgroup of $S_4$.

The only groups of order $6$ are the cyclic one and $S_3$.



Why do we have to find more than 1 element of order 2 in $S_4/N$ to show that $S_4/N$ is not cyclic?

Cyclic groups have exactly one subgroup of order $d$, for each divisor $d$ of the order of the group. In particular, they have only one subgroup of order $2$ if the order of the group is even, and thus only one element of order $2$. For a cyclic group $\langle a\rangle$ where $a$ has order $6$, this element is $a^3$.

To apply this formula at the calculation of the order of $A_4/N$ do we have to show that $N$ is a normal subgroup of $A_4$ ? (Wondering)

Yes, and no-the number of cosets (right or left) of $N$ in $A_4$ is still $|A_4|/|N|$ whether or not $N$ is normal, but $A_4/N$ is only a GROUP (that is, coset multiplication is a well-defined binary operation on cosets) if $N$ is normal.

Fortunately for you, there is nothing to prove, because of this:

If $H \lhd G$, then $H \lhd K$ for every subgroup $K$ of $G$ containing $H$.

Proof:

Since $H \lhd G$, we have $gHg^{-1} \subseteq H$ for ANY $g \in G$. If $k \in K$, then since $K$ is a subgroup of $G$, then $k \in G$, so $kHk^{-1} \subseteq H$. Since $K$ contains $H$, $H$ is thus a normal subgroup of $K$.

 

[mathmari]

Cyclic groups have exactly one subgroup of order $d$, for each divisor $d$ of the order of the group. In particular, they have only one subgroup of order $2$ if the order of the group is even, and thus only one element of order $2$.

I understand! (Yes)

find more than 1 element of order 2 in $S_4/N$

Are the elments of $S_4/N$ the 3-cycles of $S_4$ ? (Wondering)

 

[Deveno]

I understand! (Yes)
Are the elments of $S_4/N$ the 3-cycles of $S_4$ ? (Wondering)

No, the elements of $S_4/N$ are not even elements of $S_4$, they are cosets of $N$, that is to say equivalence classes of elements of $S_4$ under the equivalence relation:

$\sigma_1 \sim \sigma_2 \iff \sigma_2^{-1}\sigma_1 \in N$.

Let's enumerate these cosets, explicitly.

The first coset is, of course, $N$ itself:

$N = eN = \{e,(1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}$.

Since $(1\ 2) \not\in N$, this is another distinct coset:

$(1\ 2)N = \{(1\ 2), (3\ 4), (1\ 3\ 2\ 4), (1\ 4\ 2\ 3)\}$.

Since $(1\ 3) \not\in N, (1\ 2)N$, this gives us a 3rd coset:

$(1\ 3)N = \{(1\ 3), (1\ 2\ 3\ 4), (2\ 4), (1\ 4\ 3\ 2)\}$.

None of these three cosets contain $(2\ 3)$, giving us a 4th coset:

$(2\ 3)N = \{(2\ 3), (1\ 3\ 4\ 2), (1\ 2\ 4\ 3), (1\ 4)\}$

Note none of our cosets yet contain a $3$-cycle, so:

$(1\ 2\ 3)N$ is a 5th coset:

$(1\ 2\ 3)N = \{(1\ 2\ 3), (1\ 3\ 4), (2\ 4\ 3), (1\ 4\ 2)\}$

The last coset consists of those elements of $S_4$ we haven't listed yet:

$(1\ 3\ 2)N = \{(1\ 3\ 2), (2\ 3\ 4), (1\ 2\ 4), (1\ 4\ 3)\}$.

This particular set of "representatives" (the elements we put to the left of $N$) makes it clear $S_4/N \cong S_3$.

 

[mathmari]

No, the elements of $S_4/N$ are not even elements of $S_4$, they are cosets of $N$, that is to say equivalence classes of elements of $S_4$ under the equivalence relation:

$\sigma_1 \sim \sigma_2 \iff \sigma_2^{-1}\sigma_1 \in N$.

Ah ok... (Thinking)

Since $(1\ 2) \not\in N$, this is another distinct coset:

$(1\ 2)N = \{(1\ 2), (3\ 4), (1\ 3\ 2\ 4), (1\ 4\ 2\ 3)\}$.

Why does the element $(1 \ 2 )(3 \ 4)$ of $N$ become $(3 4)$ at $(1\ 2)N$ ? (Wondering)

 

[I like Serena]

Why does the element $(1 \ 2 )(3 \ 4)$ of $N$ become $(3 4)$ at $(1\ 2)N$ ? (Wondering)

Erm... isn't $(1 2) \circ (1 2)(3 4) = (3 4)$? (Wondering)

 

[mathmari]

Erm... isn't $(1 2) \circ (1 2)(3 4) = (3 4)$? (Wondering)

Why? I got stuck right now... (Wondering) (Sweating)

 

[I like Serena]

Why? I got stuck right now... (Wondering) (Sweating)

Where are you stuck? How would you evaluate $(12)\circ (12)(34)$? (Wondering)

Those permutations bring $1$ to $2$ and back and separately from that, it brings $3$ to $4$.

 

[Deveno]

$(1\ 2)(3\ 4)$ is this mapping:

$1 \to 2$
$2 \to 1$
$3 \to 4$
$4 \to 3$

$(1\ 2)(1\ 2)(3\ 4)$ is this mapping:

$1 \to 2 \to 1$
$2 \to 1 \to 2$
$3 \to 4 \to 4$
$4 \to 3 \to 3$, which "collapses" to:$1 \to 1$
$2 \to 2$
$3 \to 4$
$4 \to 3$,

which is the mapping $(3\ 4)$.

Tl, dr; version: $(1\ 2)$ is its own inverse.

 

[mathmari]

Those permutations bring $1$ to $2$ and back and separately from that, it brings $3$ to $4$.

$(1\ 2)(3\ 4)$ is this mapping:

$1 \to 2$
$2 \to 1$
$3 \to 4$
$4 \to 3$

$(1\ 2)(1\ 2)(3\ 4)$ is this mapping:

$1 \to 2 \to 1$
$2 \to 1 \to 2$
$3 \to 4 \to 4$
$4 \to 3 \to 3$, which "collapses" to:$1 \to 1$
$2 \to 2$
$3 \to 4$
$4 \to 3$,

which is the mapping $(3\ 4)$.

Tl, dr; version: $(1\ 2)$ is its own inverse.

I understand! Thanks for explaining! (Smile)

This particular set of "representatives" (the elements we put to the left of $N$) makes it clear $S_4/N \cong S_3$.

Do we not look at the set of the elements that the cosets contain, just at the element $g$ of the cosets $gN$ ? Why? (Wondering)

 

[Deveno]

I understand! Thanks for explaining! (Smile)



Do we not look at the set of the elements that the cosets contain, just at the element $g$ of the cosets $gN$ ? Why? (Wondering)

Yes we look at the cosets, the equivalence class of $g$, for example is $[g] = gN$.

You may find it educational to convince yourself that if $g_1 \in gN$, and $g_2 \in hN$, that $g_1g_2 \in (gh)N$.

*That* is why we usually just look at "representatives".

 

[mathmari]

No, the elements of $S_4/N$ are not even elements of $S_4$, they are cosets of $N$, that is to say equivalence classes of elements of $S_4$ under the equivalence relation:

$\sigma_1 \sim \sigma_2 \iff \sigma_2^{-1}\sigma_1 \in N$.

Let's enumerate these cosets, explicitly.

The first coset is, of course, $N$ itself:

$N = eN = \{e,(1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}$.

Since $(1\ 2) \not\in N$, this is another distinct coset:

$(1\ 2)N = \{(1\ 2), (3\ 4), (1\ 3\ 2\ 4), (1\ 4\ 2\ 3)\}$.

Since $(1\ 3) \not\in N, (1\ 2)N$, this gives us a 3rd coset:

$(1\ 3)N = \{(1\ 3), (1\ 2\ 3\ 4), (2\ 4), (1\ 4\ 3\ 2)\}$.

None of these three cosets contain $(2\ 3)$, giving us a 4th coset:

$(2\ 3)N = \{(2\ 3), (1\ 3\ 4\ 2), (1\ 2\ 4\ 3), (1\ 4)\}$

Note none of our cosets yet contain a $3$-cycle, so:

$(1\ 2\ 3)N$ is a 5th coset:

$(1\ 2\ 3)N = \{(1\ 2\ 3), (1\ 3\ 4), (2\ 4\ 3), (1\ 4\ 2)\}$

The last coset consists of those elements of $S_4$ we haven't listed yet:

$(1\ 3\ 2)N = \{(1\ 3\ 2), (2\ 3\ 4), (1\ 2\ 4), (1\ 4\ 3)\}$.

This particular set of "representatives" (the elements we put to the left of $N$) makes it clear $S_4/N \cong S_3$.

If we took at the beginning for example the coset $(14)N$, would it also be $S_4/N \cong S_3$ ? (Wondering)

 

[Deveno]

If we took at the beginning for example the coset $(14)N$, would it also be $S_4/N \cong S_3$ ? (Wondering)

Why don't you try using different "representatives" and make a Cayley table? Let me know your observations.

 

[mathmari]

$S_4/N$ is of order 6. Since there are only two groups of order 6 (cyclic and $S_3$), you just have to show $S_4/N$ is not cyclic. Hint: find more than 1 element of order 2 in $S_4/N$.

Do we have to find all the cosets, as in post #18, to show that in $S_4/N$ there are more than $1$ element of order $2$ ? (Wondering)