Proving there is no isomorphism between $S_\infty$ and $A_\infty$ My teacher said this is the case because $A_\infty$ is generated by elements of order $3$ (the 3-cycles), and $S_\infty$ is not. I understand that the 3-cycles do not generate $S_n$, and that $\phi(x) = y \implies \text{order}(x) = \text{order}(y)$, but why can't there be an isomorphism between another set of generators of $A_\infty$ (say the two cycles), and a set of generators of $S_\infty$?
 A: Suppose that $\phi:A_\infty \to S_\infty$ is an isomorphism. We know that $A_\infty = \langle X \rangle$, where $X$ is the set of $3$-cycles. It follows that $S_\infty = \phi(\langle X \rangle) = \langle \{\phi(x) : x \in X \} \rangle$.
Since $\phi(x)$ has order $3$ for all $x \in X$, it follows that $S_\infty$ is generated by a set of elements of order $3$. But every such element in $S_\infty$ is an even permutation, so this is a contradiction. Hence there can be no such isomorphism.
A: Every element in $A_{\infty}$ is a product of an even number of transpositions. If $\tau_1$, $\tau_2$, $\tau$  are transpositions, we have 
$\tau_1 \tau_2 = (\tau_1 \tau)( \tau \tau_2)$. Hence every element in $A_{\infty}$ can be written as a product of an even number of elements of the form $\eta_1 \eta_2$, where $\eta_1$, $\eta_2$ are transpositions with disjoint support. Note also that all such elements are conjugate in $A_{\infty}$ to $(1,2)(3,4)$.
Let $f\colon A_{\infty}\to S_{\infty}$ be a morphism of groups. Every element in $f(A_{\infty})$ is a product of an even number of conjugates of $f((1,2)(3,4))$, so of signature $+1$. Hence $\phi(A_{\infty})\subset A_{\infty}$. 
