Does there exist an injective homomorphism $S_6 \to S_5 \times S_5?$

Question

"Let $S_n$ denote the symmetric group on $n$ elements. Does there exist an injective homomorphism $f:S_6 \to S_5 \times S_5$ ?"

This is a problem in the past Qualifying exam in my university. I think it is easy, but too easy to be true. So we have an element of order $6$ in $S_6$, called $a$, so the image can be of order 2 or 3 if we want injectivity. However, then $f(a^2)=f(e)$ or $f(a^3)=f(e)$, which implies non-injectivity. We are done.

Clearly I thought there would be no element of order 6 in $S_5 x S_5$. So do you guys have any approach?

Answer 1

Composing a homomorphism $S_6\to S_5\times S_5$ with the two projection maps $S_5\times S_5\to S_5$ gives two homomorphisms $\phi_1$, $\phi_2:S_6\to S_5$. But $S_6$ has the simple subgroup $A_6$ which is bigger than $S_5$. So each $\phi_j$ is trivial on $A_6$ which implies that $A_6$ is in the kernel of the original homomorphism.

Answer 2

The phrase "so the image can be of order 2 or 3 if we want injectivity" is ambiguous. You should try to make that step clearer, and depending on what you meant, support it with some argument.