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Let $K$ be the subgroup $K=\{e,(1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)\}$ in $S_4$.

  1. show that $K$ is normal in $S_4$.

  2. show that $S_4/K \cong S_3$.

I know that I could try to prove this directly by the definition of normality, but that would be too long. I am thinking something like applying the First Isomorphism Theorem, like if I could some how make K a kernel of some surjective function. But I just don't know how to deal with functions of cycles.

Can anyone help please? Thanks

verticese
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  • What have you tried? If you tell us this then we will be better able to help you. And it helps us feel that we are not just doing your homework for you. For example, do you know the definition of normality? And how have you tried to apply it? Also, have you convinced yourself that $K$ is, in fact, a subgroup (it is always a good idea to check, just in case!)? – user1729 Nov 12 '14 at 15:55
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    Try to define a map $\varphi:S_4\longrightarrow S_3$ which will be the candidate for the isomorphism, if you choose it right the kernel will be $K$ and $\varphi$ will be surjective, then it suffices applying the first isomorphism theorem to get $(b)$.. – PtF Nov 12 '14 at 15:59
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    @verticese You should edit this into your question then - the current answer is just using this definition of normality. You won't get the answer you are looking for unless you ask for it! :-) – user1729 Nov 12 '14 at 16:05
  • As a general rule, since you're already being asked to show that the quotient group is isomorphic to something else, you might as well just go straight to that with the isomorphism theorems...that gives you normality for free, and saves you time/effort – Alan Nov 12 '14 at 16:32

2 Answers2

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Let $g \in S_4$ and $k \in K$. We need to show $g^{-1}kg \in K$. Recall that conjugating $k$ by $g$ gives a permutation that has the same cycle structure as $k$. Observe that $K$ contains all permutations of cycle structure $(i,j)(k,\ell)$ and $(i)(j)(k)(\ell)$, hence $g^{-1}kg \in K$ for all $g \in S_4$. Thus $K$ is normal in $S_4$.

Consider the quotient group $S_4/K$. We need to show it is isomorphic to $S_3$. The quotient group has order 6, and up to isomorphism there are only two groups of order 6, namely $S_3$ and $C_6$. But if a coset $gK$ in the quotient group has order 6 for some $g \in S_4$, then $g^6 \in K$, but $S_4$ does not have any nonidentity permutation whose sixth power is in $K$ and whose lesser powers are not in $K$.

Ashwin Ganesan
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    This doesn't answer the question as it is currently stated, about trying to find a surjection $S_4\rightarrow S_3$ with the given subgroup as kernel... – user1729 Nov 12 '14 at 16:22
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Hint:

For 1, note that $K$ contains all cycles of the form $(a,b)(c,d)$ in $S_4$

What can you say about conjugation of such a cycle? How can this help you?

For 2 $|S_4|/|K|=6$.

So $S_4/K\cong \mathbb{Z}_6\,\,\,\,\,$ or $\,\,\,\,S_4/K\cong S_3$.

Why can you eliminate the possibility of $\mathbb{Z}_6$?

Swapnil Tripathi
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