# Subgroups of $S_4$ and homomorphisms

Good day,

I am given a problem: there is a homomorphism from $S_4$ onto $Z_2,$ and I am asked to find its kernel. By the first isomorphism theorem, I know that the order of the kernel must be 12. I know that $S_4$ has a subgroup of order 12, namely $A_4,$ the group of even permutations. I am tempted to say that it is the kernel. But the following concerns me:

1) How do I know that it is the subgroup of order 12 of $S_4$? How do I know there are no others? I know odd permutations do not form one, but is there a simpler/more elegant way to say it rather than going into "consider a group with some even, some odd permutation cycles. Let's show it's not closed and thus can't be a subgroup of $S_4$?"

2) A more general question, if a group has a subgroup, its order divides the group's order. But how many subgroups of such order can it maximally have (not talking about cyclic groups, who have unique subgroups for every order)? That is, let's say if I have a group of order 24, it could have 2 disjoint (besides the identity) subgroups of order 12 presumably, but could it have more? Obviously, with elements of both of the aforementioned disjoint (besides the identity) subgroups. What is the general theory behind this that could be used to answer the question (I feel like I either haven't studied it yet, or am missing something important). I know when we are dealing with cosets, everything gets broken down into equal-sized chunks, so if order of a subgroup is k and of group is x, you have exactly $\frac{x}{k}$ cosets. Thank you very very much.

-
Note that no two subgroups can be disjoint! They share the identity definitely! – user21436 Jan 28 '12 at 20:27
@KannappanSampath: corrected thank you. – algebraist_not Jan 28 '12 at 20:28
Do you know the abelianization of $S_4$? – Qiaochu Yuan Jan 28 '12 at 20:36
@QiaochuYuan: I am afraid we haven't studied that yet. – algebraist_not Jan 28 '12 at 20:38
It is, perhaps, worth pointing out that it is possible for a group to have more than one subgroup of index $2$. For example, a Dihedral group of order $2^n$ has $3$ subgroups of index $2$. (This is because then $G$ is a $2$-group of maximal class (which means it has a lower central series of length $n-1$) and so it necessarily has $3$ subgroups of index $2$. More generally, replace the $2$ with an arbitrary prime $p$ and such a group has $p+1$ subgroups of index $p$...a proof of this should be in Robinson's book "Topics in the Theory of Groups". It's the Burnside Basis Theorem.) – user1729 Jan 28 '12 at 22:04

I have the following recipe:

• Prove that the only non-trivial normal subgroups of $S_4$ are $A_4$ and a group isomorphic to Klein $4$-Group.

• Deduce from here that, $A_4$ can be the only group of order $12$.

Alternatively, this can be thought of by replacing $\mathbb Z_2$ with the multiplicative cyclic group of order $2$ and noting that signature is a homomorphism!

Your Question (As I understand it!)

Can there be more subgroups $H$ of cardinality $|H|$ than $\dfrac{|G|}{|H|}$? (Excerpted from your comment!)

Answer: Consider the Klein $4$-Group. And look at Subgroupsof order $2$. There are three of them, one more than $2$.

-
Thank you, I will try that. Now about my second question, can a group of order for example 24 have three subgroups of order 12? Nothing magical about 24, just wondering if a group $G$ can have more subgroups of order $k$ than $|G|/k.$ – algebraist_not Jan 28 '12 at 20:49
@algebraist_not I have added an answer. Look at that leave comments for me! – user21436 Jan 28 '12 at 20:57
Thank you this is good and very helpful to me. – algebraist_not Jan 28 '12 at 20:57

Because $S_4$ has order $24$, any subgroup $H$ of order $12$ has index $2$ and is thus normal.

Therefore $H$ is an union of conjugacy classes of $S_4$. There are $5$ conjugacy classes which are determined by cycle structure. Since $H$ is a subgroup, it has to contain the conjugacy class of the identity. Now when we look at the number of elements in the other conjugacy classes, there is only way to get $12$ elements. This is by taking the three conjugacy classes of $A_4$.

-

1) I might as well explain my comment. Any group $G$ has a subgroup $[G, G]$, the commutator subgroup of $G$, generated by elements of the form $aba^{-1} b^{-1}$. The significance of the commutator subgroup is that any homomorphism $\phi : G \to A$ from $G$ into an abelian group has the property that $\text{ker}(\phi)$ contains $[G, G]$. In fact, the quotient $G \to G/[G, G]$ is universal with this property; that is, any other homomorphism $\phi : G \to A$ from $G$ into an abelian group factors through a $G/[G, G] \to A$. Thus we call $G/[G, G]$ the abelianization of $G$; it is the universal way to map $G$ into an abelian group.

One can verify by direct computation that $[S_4, S_4] = A_4$, and it is true more generally that $[S_n, S_n] = A_n$, so the abelianization map $S_n \to S_n/A_n \cong \mathbb{Z}/2\mathbb{Z}$ is just the sign homomorphism. Hence there is a unique nonzero map from $S_n$ to $\mathbb{Z}/2\mathbb{Z}$, the kernel of which is the unique subgroup of index $2$.

2) A lot. For example, let $G = (\mathbb{Z}/2\mathbb{Z})^n$ be an $n$-dimensional vector space over the finite field $\mathbb{F}_2$. The only subgroups have order $2^k$, they are precisely the $k$-dimensional subspaces, and there are $${n \choose k}_2 = \frac{[n]_2 [n-1]_2 ... [n-k+1]_2}{[k]_2 [k-1]_2 ... [1]_2}$$

of them, where $[n]_2 = 2^n - 1$ (see $q$-binomial coefficient). In particular, there are $2^n - 1$ subgroups of order $2$ since each nonzero element of $G$ generates a unique such subgroup.

In general, the answer to the question "how many subgroups of order $k$ does a given group $G$ have?" depends very much on the prime factorization of $k$ and on the properties of $G$. For example, if $k = p$ is prime, any non-identity element of a subgroup of order $p$ generates it, hence the intersection of any pair of distinct such subgroups is necessarily just the identity. It follows that $G$ has at most $\frac{|G|-1}{p-1}$ such subgroups, and the example of $G = (\mathbb{Z}/p\mathbb{Z})^n$ shows that this is tight.

Since a subgroup of order $k$ is in particular a subset of the non-identity elements of $G$ of size $k-1$ (together with the identity), then a trivial upper bound is that $G$ has at most ${|G|-1 \choose k-1}$ subgroups of order $k$, but I don't expect this to ever be tight for $k > 2$.

If $k = p^n$ for $p$ a prime, and furthermore $p^n$ is the greatest power of $p$ dividing $|G|$, then the number of subgroups of order $k$ is controlled by the Sylow theorems; in particular, it divides $\frac{|G|}{k}$.

-

As it happens, less than two weeks ago I gave a talk in my local number theory seminar ("local" modifies "seminar", not "number theory"!) in which I chose to give a detailed proof of why for all $n \geq 2$, $S_n$ has a unique nontrivial homomorphism to $\{ \pm 1 \}$. This was a strange thing to do for that audience, and I can't remember why I did it...but it will come in handy now!

Here are two facts about $S_n$ that everyone who knows what $S_n$ is should know:

1) Two elements are conjugate in $S_n$ iff they have the same cycle type.
2) $S_n$ is generated by two-cycles. Indeed, it is certainly generated by all of its cycles, and a $k$-cycle can be written as a product of $2$-cycles:

$(a_1,a_2,\ldots,a_k) = (a_{k-1}, a_k) \cdots (a_2, a_3) (a_1, a_2)$.

Now let $f$ be any homomorphism from $S_n$ into a commutative group $G$. Since $S_n$ is generated by two-cycles, $f$ is determined by where it maps every two-cycle. But moreover, since all two-cycles are conjugate in $S_n$, their images under $f$ will be conjugate in the abelian group $G$, and thus equal. Thus there are, at most, as many homorphisms from $S_n$ to $G$ as there are elements $x \in G$ with $x^2 = 1$. When $G = \{ \pm 1 \}$ there are therefore at most two homomorphisms: the trivial one, and the one which sends every two-cycle to $-1$.

One can further check that there is a group homomorphism $S_n \rightarrow \{ \pm 1\}$ sending every two-cycle to $-1$ by considering the map $\iota: S_n \rightarrow \operatorname{Aut}_{\mathbb{C}}(\mathbb{C}^n)$ which sends a permutation $\sigma$ of $\{1,\ldots,n\}$ to the unique linear transformation sending the standard basis $e_1,\ldots,e_n$ to the permuted basis $e_{\sigma(1)},\ldots,e_{\sigma(n)}$. By easy linear algebra every such matrix has determinant $\pm 1$, and the determinant of any two-cycle is the elemtentary matrix obtained by switching two rows of $I_n$ so has determinant $-1$. Thus $\operatorname{det} \circ \iota: S_n \rightarrow \{ \pm 1\}$ is a surjective group homomorphism, usually called the sign homomorphism. Its kernel, usually called $A_n$, is an index $2$ subgroup of $S_n$ and is the unique such, since a different index $2$ subgroup of $S_n$ would give rise to a different surjective homomorphism from $S_n$ to $\{ \pm 1\}$.

Remark: If I am not mistaken, the above reasoning is enough for a complete proof that $[S_n,S_n] = A_n$ for all $n \geq 2$.

-
I am surprised you supply such fine details in a Seminar. How I wish that those speakers in the talks I attend supply details like this. I am always lost at the 15th minute in any talk. Given that I am an undergrad in my very first year, people blame me for attending talks! I should probably tell them this! – user21436 Jan 30 '12 at 0:23