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I have to solve these exercises but I don't know how to do:

  • Describe (if they exist) all subgroups of order 6 in $S_4$;

  • Describe (if they exist) all subgroups of order 8, 10, 12, 15, 20, 24 in $S_5$.

I was asked to solve them after my teacher explained Sylow theory, and semidirect products. Is there a general method to do this kind of exercise?

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    $\begingroup$ Yes. Classify all isomorphism types of groups of a given order. Find a nice presentation of them using generators and relations. Find a set of permutations satisfying said relations (and no extra). $\endgroup$ – arctic tern Oct 23 '16 at 16:12
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    $\begingroup$ First question see here and similar duplicates. Second question see here. For "general method" see also here. $\endgroup$ – Dietrich Burde Oct 23 '16 at 16:16
  • $\begingroup$ @arctic tern, Sorry, can you make an example? What would you do to describe all subgroups of order 10 in $S_5$? $\endgroup$ – GRE Oct 23 '16 at 17:36
  • $\begingroup$ @GRE Okay. The two groups of order $10$ up to isomorphism are $C_5\times C_2\cong C_{10}$ and $C_5\rtimes C_2\cong D_{10}$. In the first case, we would need two commuting permutations of orders $5$ and $2$ respectively. One may classify all permutations of those orders in $S_5$ and see that no pair commutes, so that rules that out. For the second case, we need the same kind of pair but instead of commuting, conjugating the order $5$ permutations by the order $2$ one yields its inverse. One may find all such pairs. $\endgroup$ – arctic tern Oct 23 '16 at 18:22
  • $\begingroup$ @arctic tern Okay, it seems to me a condition to approach this problem is to know who are all the groups of the requested order up to isomorphism, and then, I have to make some considerations about the elements to be able to exclude the wrong possibilities, is it correct? In this case, I have to find this pair of permutations to conclude $D_{10} $ is a subgroup: this work is "manual"? I mean, there is no way to use, for example, Sylow theory? And, last question related to the first, is there a method to find all these subgroups without the classification of the groups of the same order? $\endgroup$ – GRE Oct 23 '16 at 21:52
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The isomorphism classes of subgroups of $S_5$ of order $8\le n\le 24$ are given by the groups $$ D_4, D_5,C_2\times S_3, A_4, Aff(5), S_4 $$ of order $$ 8,10,12,12,20,24 $$ respectively. In particular, $S_5$ has no subgroup of order $15$. A proof goes as follows (for the last statement):

Proof by contradiction. Suppose $H$ is a subgroup of $S_5$ of order $15$. Then (by Lagrange’s theorem), each element of $H \setminus \{\sigma_{id}\}$ is either a $3$-cycle or $5$-cycle. Suppose there are $m$ subgroups of order $3$ and $n$ subgroups of order $5$. Then $H$ has $2m$ elements of order $3$ and $4m$ elements of order $5$. Now since each element other than the identity has order $3$ or $5$, we have that $14 = 2m+4n$, and $7 = m+2n$. Thus $m\in\{1,3,5,7\}$ and $n\in\{0,1,2,3\}$. For $i=1,2,3,4,5$,let $H_i =\{σ\in H \mid σ(i)=i\}$, that is, the set of permutations $σ \in H$ which fix $i$. For $i,j \in \{1,2,3,4,5\}$, $i\neq j$, let $H_{ij} = H_i ∩ H_j$. Then we see that for all $i,j$, $H_{ij }=<\sigma>$ for some 3-cycle $σ$. Note that $H_i$ and $H_{ij}$ are also subgroups of $H$, and the (non-trivial) $H_{ij}$’s are exactly the subgroups of order $3$.

Suppose $m = 7$. Then, each element $H \setminus \{\sigma_{id}\}$ is a $3$-cycle. Since there are $7$ nontrivial subgroups $H_{ij}$, we can find non-trivial $H_{i_1j_1}$ and $H_{i_2j_2}$ where $\{i_1,j_1\}∩\{i_2,j_2\} = \emptyset$. Let $σ\in H_{i_1j_1}$ and $τ\in H_{i_2j_2}$. Then $γ=τσ\in H$ but $γ$ does not fix $i_1,j_1,i_2$, or $j_2$. Thus $γ$ cannot be a $3$-cycle, a contradiction. We conclude that $m < 7$ and thus $n > 0$. Suppose $m ≥ 3$. Then (by simple counting), there exists non-trivial $H_{ij}$ and $H_{ij'}$ . Let $σ \in H_{ij}\setminus \{σ_{ij}\}$ and $τ \in H_{ij′} \setminus\{σ_{id}\}$. Then $γ = τσ \in H_{ij′′}$, where $j′′\neq j,j′$. This implies that $|H_i| ≥ 1 + 3 · 2 = 7$. However, since $|H_i|$ divides $|H| = 15$, it follows that $|H_i| = 15$. But this would imply that all elements of $H$ are 3-cycles, a contradiction.

We conclude from the above that $m=1$ and hence $n=3$.Let $τ_i$, $i=1,2,3$ be three $5$-cycles in $H$ where $τ_i\not\in < τj >$ for all $i\neq j$. Let $σ$ be a $3$-cycle in $H$. For $i = 1,2,3$, let $γ_i = στ_1^i$. We see that $γ_1\not\in< σ >$, and $γ_1\not\in < τ_1 > $. Without loss of generality, we can assume $γ_1 \in <τ_2 >$.Then $γ_2$ does not belong to $<σ>,<τ_1 >$, or $<τ_2 >$. Thus $γ_2 \in<τ_3 >$. However,one now sees that $γ_3$ does not belong to any of $<σ>$ or $< τ_i >$, $i = 1, 2, 3$. This gives a final contradiction.

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