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Let $G = \{\sigma \in S_n\mid \epsilon(\sigma) = 1\}$ with some $n\in\mathbb{N}$, $\ast$ is the multiplication (i.e. composition) of permutations. Is $(G,\ast)$ a group?

I have some questions when proving it. For example, first of all I have to prove that $\ast$ is binary but how can I show that

if $\epsilon(\sigma)=1$ and $\epsilon(\tau)=1$ then $\epsilon(\sigma\tau)=1$?

We know that it's associative because the composition of permutations is, and the identity element is the identity permutation (as it belongs to $G$, $\epsilon(id)=1$) but what about the inverse? The inverse of a permutation in $G$ will be $\tau=\sigma^(-1)$ but how can I prove that it belongs to $G$?

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    $\begingroup$ You haven't defined $\epsilon$. $\endgroup$ – Derek Holt Jan 29 '15 at 9:51
  • $\begingroup$ @DerekHolt: I think here $\epsilon$ means sign of permutation. Some authors (e.g. Dummit & Foote) used this notation. But agree with you that OP should have mentioned it in question. $\endgroup$ – Krish Jan 29 '15 at 10:05
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    $\begingroup$ Well yes, but the question was really aimed at the OP! $\endgroup$ – Derek Holt Jan 29 '15 at 14:01
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I presume that $\epsilon$ is the sign of a permutation, and that you know already that this is well-defined(?)

Suppose, then, that $\sigma$ and $\tau$ have sign equal to $1$. This means one can write each of them as a product of an even number of transpositions; let $k$ (resp. $\ell$) be the number of transpositions in some representation of $\sigma$ (resp. $\tau$). Then the product $\sigma\tau$ can be written as a product of $k+\ell$ transpositions. Since $k$ and $\ell$ are even, their sum is even, and hence the product $\sigma\tau$ has sign equal to $1$.

The proof for inverses is similar: use the fact that if you write $\tau$ as a product of transpositions, then $\tau^{-1}$ can be written as a product of the same transpositions, but in reverse order.

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