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Show that if $H$ is a subgroup of $S_n$ the symmetric group of order $n$, then either every member of $H$ is an even permutation or exactly half of the members are even.

I can see that if $a,b$ are arbitrary even elements in $H$, then their product and any combination of their products will be even, which implies $H$ is full of even permutations, but I'm not sure how to show the half-even half-odd part.

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    $\begingroup$ Say $o$ is an odd element of $H$. Think about the map $h\mapsto oh$. $\endgroup$ Jul 21, 2016 at 15:33
  • $\begingroup$ I'm unfamiliar with that mapping notation, could you explain what that arrow means? $\endgroup$
    – Oliver G
    Jul 21, 2016 at 15:35
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    $\begingroup$ Define a map $f:H\to H$ by $f(h)=oh$. $\endgroup$ Jul 21, 2016 at 15:36
  • $\begingroup$ It seems it would be a bijective map from even functions to odd functions, which would imply the same cardinality. Does that make sense? $\endgroup$
    – Oliver G
    Jul 21, 2016 at 15:39
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    $\begingroup$ Makes sense to me. $\endgroup$ Jul 21, 2016 at 15:40

4 Answers 4

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Let $H$ be a subgroup of $S_n$. If $H$ contains no odd permutations, then $H$ contains only even permutations, and we're done.

Otherwise, let $o\in H$ be an odd permutation and consider the function $f:H\rightarrow H$ that multiplies each element by $o$. Note that this function is bijective: it's injective, with an inverse function that multiplies each element by $o^{-1}$, and it's surjective, because for every element $h\in H$, we can find an element $h\cdot o^{-1}$ that $f$ maps into it.

Note that multiplying by an odd permutation changes odd permutations into even permutations and vice versa. It follows that $f$ is a bijection that perfectly pairs up the odd permutations in $H$ with the even permutations. Hence there are exactly as many odd permutations as even permutations in $H$.

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Consider the homomorphism:

$\text{sgn}: S_n \to \{-1,1\}$ (with kernel $A_n$).

When restricted to $H$, this gives a homomorphism $H \to \{-1,1\}$.

We have two possibilities:

1.) $\text{sgn}(H) = \{1\}$, (that is $H \subseteq A_n$), or:

2.) $\text{sgn}(H) = \{-1,1\}$. Use the first isomorphism theorem, here.

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Cosets are equal in size. If there exist odd elements in $H$, then the set of odd elements $H_{\mathrm{odd}}$ is a coset of the subgroup of even elements $H_{\mathrm{even}}$. Prove this! Can you proceed from there?

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Let $H$ be a subgroup of $S_n$

If every member of $H$ is even permutation, then $H\subseteq A_n$, the subgroup of $S_n$ consisting of all even permutations.

Suppose not, i.e. there is at least one element namely $\sigma_1$ which is an odd permutation. Let $m_e,m_o$ denote number of even, odd permutations in $H$. If $\sigma_1,\sigma_2,\ldots,\sigma_{m_o\ }$ are odd cycles, then $\sigma_1^2,\sigma_1\sigma_2,\ldots,\sigma_1\ \sigma_{m_o}$ are all distinct even permutations, so there are at least that many even permutations as many there are odd ones, which implies $m_e\geq m_o$.

On the other hand, if $\alpha_1,\ldots,\alpha_{m_e\ }$ are even permutations, then $\sigma_1\ \alpha_1,\ldots,\sigma_1\ \alpha_{m_e\ }$ are all odd permutations, so there are at least that many odd permutations as many there are even ones, which gives $m_o\geq m_e$.

Thus, $m_e=m_o=|H|/2$

The result we have proved is: If $H$ is a subgroup of $S_n$, then either $H\subseteq A_n$ or $|H\cap A_n|=|H|/2$

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