Limitations on the order of a subgroup of $S_5$

Question

The goal is to show $S_5$ has no subgroup of order $15$ or $30$, but does have one of order $20$. I think I have part of the first part of it:

There are only two groups of order $15$, $\mathbb{Z}_3\times \mathbb{Z}_5$ and $\mathbb{Z}_{15}$ (Of course, these two groups are isomorphic). While $S_5$ contains copies of $\mathbb{Z}_3$ and $\mathbb{Z}_5$, neither of them are normal, and hence their direct product is not a subgroup. Alternatively, I'm sure $S_5$ doesn't have an element of order $15$, but I forget how to show it, which would lead to $S_5$ not containing $\mathbb{Z}_{15}$, therefore no subgroup of order 15.

I know that if a subgroup of $S_5$ has index $k\le n$, then the only possibilities for $k$ are $1,2,5$. Thus, as an order $30$ subgroup would have index $4$, such a subgroup doesn't exist.

I'm then having trouble showing there's a group of order $20$. I mean, I've seen on the groupprops wiki what it's generated by, but I'm not sure on how I'd come up with elements to test (I suppose since $4\cdot 5=20$, but that logic doesn't work with $3\cdot 5=15$). Any guidance would be appreciated.

Answer 1

Consider the Sylow $5$-subgroups. It's easy to verify that there are six of these. In general, if $n_p$ denotes the number of Sylow $p$-subgroups and $P$ is any of these subgroups, then $n_p = |G:N_G(P)|,$ where $N_G(P)$ is the normalizer of $P$.

So in this example, if $P$ is any Sylow $5$-subgroup, then we have $|G:N_G(P)| = 6$, hence $|N_G(P)| = |G|/6 = 120/6 = 20$.

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Suppose that we want to find an explicit subgroup of order $20$. By Cauchy's theorem, such a subgroup must contain an element of order $5$, and in $S_5$ the only such elements are $5$-cycles, so let's start with some arbitrary $5$-cycle $(abcde)$.

Let $H$ denote the subgroup generated by $(abcde)$, hence $H$ has order $5$.

Let's try to find some element which is not in $H$ but which normalizes $H$. For example, let's find some $k \in G = S_5$ such that $k(abcde)k^{-1} = (abcde)^2 = (acebd)$. It's easy to check that $k = (bced)$ satisfies this requirement. Since conjugation by $k$ sends $(abcde)$ to a power of $(abcde)$ and both of these generate $H$, it follows that $k$ normalizes $H$.

Now let $K$ be the subgroup generated by $k$, hence $K$ has order $4$.

Let $N = N_G(H)$ denote the normalizer of $H$ in $G$. Then $H \lhd N$ by definition. Also, we just showed that $k \in N$, hence $K = \langle k \rangle \leq N$. It follows that $HK$ is a subgroup of $N$, hence also a subgroup of $G$. The order of $HK$ is $$|HK| = |H||K|/|H \cap K| = |H||K| = 5\cdot 4 = 20$$ since $|H \cap K| = 1$ (because $|H|=5$ and $|K|=4$ are relatively prime). Since $HK$ is a subgroup with the desired order, we're done. Note that this is a semidirect product since $H \lhd HK$ but $K \not\lhd HK$.

Answer 2

As for proving that there are no subgroups of order $15$, I think it looks great.

$S_5$ cannot have a subgroup of order $15$ because all such subgroups are cyclic, and $S_5$ does not have elements of order $15$, to prove this notice that the order of an element is equal to the least common multiple of its cycle lengths, so it would have to have a cycle of length $5$ or $15$, the first case immediately forces the permutation to be a $5$-cycle and the second is impossible.

For the subgroup of order $30$, I would just use this lemma: If a group has order $2k$ with $k$ odd then the group has a a subgroup of order $k$.

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Now lets try to build the subgroup of order $20$.

It is easy to see with sylows theorems that there can be only one $5$-sylow subgroup (since the number of such subgroups must be $1\bmod 5$ and divide $4$)

because of Cauchy's theorem, without loss of generality it must contain $e$ and all the powers of $(1,2,3,4,5)$.

The subgroup must also contain an element of order $2$,call it $\gamma$, without loss of generality $\gamma$ does not move $1$. Notice that $\gamma(1,2,3,4,5)\gamma^{-1}=(1,\gamma(2),\gamma(3),\gamma(4),\gamma(5))$ must be another power of $\gamma$, and so there are only $4$ possibilities for $\gamma$ depending on the possibilities for $\gamma(1,2,3,4,5)\gamma^{-1}$.

$(1,2,3,4,5)$ implies $\gamma=e$

$(1,3,5,2,4)$ implies $\gamma=(2,3,5,4,2)$

$(1,4,2,5,3)$ implies $\gamma=(2,4,5,3)$

$(1,5,4,3,2)$ implies $\gamma=(2,5)(3,4)$.

Notice that $(2,4,5,3)^2=(2,5)(3,4)$.

The subgroup generated by $(1,2,3,4,5)$ and $(2,5)(3,4)$ is clearly isomorphic to the dihedral group on $5$ elements.

So we need to take the subgroup generated by $(1,2,3,4,5)$ and $(2,4,5,3)$.

How do we prove it has exactly $20$ elements?

Recall $(2,4,5,3)^{-1}(1,2,3,4,5)(2,4,5,3)=(1,4,2,5,3)$.

This allows us to "commute" $(2,4,3,5)$ with $(1,2,3,4,5)$, and clearly $(2,4,5,3)$. And clearly $(2,4,5,3)$ commutes with $(2,4,5,3)^2=(2,5)(3,4)$.

Doing this we can show that every element in the subgroup generated by $(2,4,5,3)$ and $(1,2,3,4,5)$ is of the form $(2,4,5,3)^k(s)$ or $s$ with $s$ inside the subgroup generated by $(2,5)(3,4)$. Since $(2,4,5,3)^2=(2,5)(3,4)$ and the order of $(2,4,5,3)$ is $4$ we can let $k=0$ or $k=1$.

Conclusion: The subgroup $\langle (2,5)(3,4), (1,2,3,4,5)\rangle$ has index at most $2$ in $\langle (2,4,5,3),(1,2,3,4,5)\rangle$. They are not equal, because one contains odd permutation. Therefore $\langle (2,4,5,3),(1,2,3,4,5)\rangle$ has exactly two times the number of elements as $\langle (2,5)(3,4),(1,2,3,4,5)\rangle$. But this second subgroup has $10$ elements, as it is isomorphic to the dihedral group.

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I also tried to find a $5$ vertex structure that had $20$ automorphisms, but failed.