Let $\tau_1, \tau_2 \in A_6$ be permutations with cycle-type $(2, 4)$. Show there exists $\tau \in A_6: \tau \tau_1 \tau^{-1} = \tau_2.$

Question

Let $\tau_1, \tau_2 \in A_6$ be permutations with cycle-type $(2, 4)$. Show there exists $$\tau \in A_6\mid \tau \tau_1 \tau^{-1} = \tau_2$$

I know $\tau_1 = (a b c d)(e f)$ and $\tau_2 = (a^{'}b^{'}c^{'}d^{'})(e^{'}f^{'})$ and I've proved that these lies in $A_6$.

I know we can write $\tau (a b c d)(e f) \tau^{-1} = (\tau(a) \tau(b) \tau(c) \tau(d))(\tau(e) \tau(f))$ and I know there exists $\tau \in S_6$ satisfying $\tau \tau_1 \tau^{-1} = \tau_2$, but how do I determine if it lies in $A_6$ without knowing the form of this permutation ?

Note: $A_6$ denote the alternating group of $S_6$ determined by the $sgn$ homomorphism.

Answer 1

If your permutations are $\tau_1 = (a\ b)(c\ d\ e\ f),$ and $\tau_2 = (x\ y)(z\ w\ u\ v),$ then $\tau$ should map $a$ to $x,$ $b$ to $y,$ $c$ to $z$ and so on.

EDIT I missed the $A_6$ part, but the point is that if $\tau,$ as above is not in $A_6,$ map $b$ to $x$ and $a$ to $y$ (and everything else as before) instead.