Prove or disprove $(1,3)A_6=A_6(1, 3)$

Question

Let $A_6$ be the alternating group of order 6. Then prove or disprove $$(1,3)A_6=A_6(1, 3).$$ If above holds, then $$(1,3)A_6(1, 3)=A_6.$$So, if we prove the last equality, then the given equality also holds. It is easy to see that $$(1,3)A_6(1, 3)\subseteq A_6.$$ How to proceed further? Please help.

Answer 1

$(1,3)A_6(1,3)\subset A_6$ since its elements are product of an even number of transpositions. The two groups are thus equal since they have the same cardinal.

Let $f(x)=(1,3)x(1,3)$. $f$ is an automorphism (the inner conjugation by $(1,3)$. $f(A_6)=(1,3)A_6(1,3)$, we deduce that $(1,3)A_6(1,3)$ has the same cardinal than $A_6$.

Answer 2

HINT: Since $(1,3) \notin A_6$ but $(1,3) \in S_6$, if we know $A_6 \unlhd S_6$ (any index $2$ subgroup is normal), we know by normal subgroup definition that $\forall g \in S_6$, $gA_6g^{-1} = A_6$.