Is every group of odd order isomorphic to a subgroup of $A_n$ for some $n$? 
Is every group of odd order isomorphic to a subgroup of $A_n$ for some $n$?

If not, what is a counterexample; if so, how can I prove it?
Hints will be appreciated.
 A: Every finite group is isomorphic to a subgroup of $S_n$. Now take $G$ a subgroup of odd order included in $S_n$. Let $g\in G\subseteq S_n$, write its unique decomposition into cycles :
$$g=c_1...c_r\text{ then } \epsilon(g)=\epsilon(c_1)...\epsilon(c_r) $$
Now you know how to compute the signature of a cycle, namely if $c$ is a cycle of length $l$ then $\epsilon(c)=(-1)^{l+1}$. 
Hence :
$$\epsilon(g)=(-1)^{(l_1+1)+...+(l_r+1)} $$
Where $l_i$ is the length of $c_i$. To finish, I claim that $l_i$ must be odd because you know that $o(c_i)=l_i$ and (all $c_i$'s are commuting) $o(g)=ppcm(o(c_1),...,o(c_r))=ppcm(l_1,...,l_r)$ so if one of the $l_i$'s were even so would be $o(g)$ and (by Lagrange's theorem) so would be $|G|$. Hence $(l_i+1)$ is always even and $\epsilon(g)=1$. This means that $G\subseteq Ker(\epsilon)=A_n$. 
A: Hint


*

*Say, left multiplication by any element $g$ of a finite group $G$ permutes the elements of $G$ (and by uniqueness of the identity element this permutation is distinct for each element $g$), so we can always regard $G$ as a subgroup of $S_G \cong S_{\# G}$.

*For any $m$ there is an injective homomorphism $S_m \hookrightarrow A_{m + 2}$.
Remark Putting these facts together gives us an upper bound on the $n$ required: Every finite group $G$ is isomorphic to a subgroup of $A_{\#G + 2}$.
