Hw7 Problem 10

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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.


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

\begin{align} \phi (\sigma_{1}) = \phi (\sigma_{2}) \end{align}


\begin{align} \lambda \sigma_{1} = \lambda \sigma_{2} \end{align}

By left cancellation,

\begin{align} \sigma_{1} = \sigma_{2} \end{align}

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

\begin{align} \phi (\lambda^{-1} \tau) = \lambda (\lambda^{-1} \tau) \end{align}
\begin{align} = (\lambda \lambda^{-1})\tau \end{align}
\begin{align} = e\tau \end{align}
\begin{align} = \tau \end{align}

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.

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