# Show that if $H$ is a subgroup of $S_n$, then either every member of $H$

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.

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

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 $$H$$.

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.

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?

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 $$|H\cap A_n|=|H|/2$$