Return to Homework 7, Glossary, Theorems

### Problem 10

(*) Show that for every subgroup $H$ of $S_{n}$ for $n \geq 2$, either all the permutations in $H$ are even or exactly half of them are even.

**Solution**

Let $n \geq 2$ and $H \leq S_{n}$. If $H$ contains all even permutations then no work must be done. Suppose $\lambda \in H$ and $\lambda$ is odd. Define $H_{E}$ as the even subset of $H$ and $H_{O}$ as the odd subset of $H$. Let $\phi : H_{E} \rightarrow H_{O}$ as $\phi (\sigma) = \lambda \sigma$ for all $\sigma \in H_{E}$. Then $\lambda \sigma$ is an odd permutation, i.e., $\lambda \sigma \in H_{O}$. Suppose

(1)Then

(2)By left cancellation,

(3)Then $\phi$ is one-to-one.

Now let $\tau \in H_{O}$. Consider the element $\sigma = \lambda^{-1} \tau$ which must be in $H$ due to group properties and must be even since $\lambda^{-1}$ and $\tau$ are odd. Then

(4)So, $\phi$ is onto.

Since $\phi : H_{E} \rightarrow H_{O}$ is a one-to-one map of $H_{E}$ onto $H_{O}$, $|H_{E}|=|H_{O}|$.

Therefore either all permutations within $H$ are even or exactly half of them are.