Prove that $C_{A_7}((567))\cong H \times A_4$ Let $H=(<567>)\subset{A_7}$. And let $C=C_{A_7}((567))$ denote the centralizer of $(567)$ in $A_7$.
Prove that $C_{A_7}((567))\cong H \times A_4$
I'm fairly certain that I can use the First Isomorphic Theorem (FIT) here if I can create/find a map between the two groups. I know there are 3 elements in $H$ and there are 24 elements in $A_4$. I know that all the elements in $A4$ are going to be disjoined with (567), and so will commute with it. 
I'm trying to picture what the centralizer looks like. I know that 3-cycles exist in $A_n$. In this case we'd need elements that commute with (567). So if I'm thinking clearly, we could create a 3-cycle with the elements $\{1,2,3,4\}$. So there are 4 choices for the first spot in the 3-cycle, 3 choices for spot 2 and 2 choices for spot 3. So there should be 24 possible choices for the 3-cycles in this centralizer. Similarity, we can consider double transpositions, $(ab)(cd)$. Using the same line of thought, 4 choices for a, 3 choices for b, etc. We conclude that there would be 24 double transpositions. I know I am missing the identity as well, but I cannot think of what the other elements look like.
Meanwhile, considering $H \times A_4$, we can clearly see 36 elements. Namely
$$(\epsilon ,A_4) \\
((567) ,A_4)\\
((567)^{-1} ,A_4)
$$
If I can create a map where it sends types to a certain set, ie 3-cycles of the centralizer could be mapped to $((567) ,A_4)$, I might be able to define an isomorphism with a trivial kernel to use the FIT.
Edit: Much of my counting here is just wrong.
 A: An alternative argument would be to count the size of the conjugacy class.
Because $A_7$ is triply transitive, all the 3-cycles fall into the same conjugacy class. How many 3-cycles are there? The set of three numbers from $\{1,2,3,4,5,6,7\}$ can be selected in ${7\choose 3}=35$ different ways, and using those three numbers we get two 3-cycles - $(abc)$ and $(acb)$. Hence there are $2\cdot35=70$ 3-cycles altogether.
But the conjugates of a given element of a group are in bijective correspondence with the coset of the centralizer of the said element, so this implies that $C=C_{A_7}((567))$ is of index $70$ in $A_7$. Therefore 
$$|C|=\frac{|A_7|}{70}=\frac{2520}{70}=36.$$
The group you have described is clearly contained in $C$, and it has $36$ elements. Therefore it must be the centralizer.
A: OP's counting argument is not correct. You can count the elements of a given cycle structure as Jyrki has or like this:
In Jyrki's answer you want to form a $3$-cycle $(a\ b \ c)$ from $\{1,2,3,4,5,6,7\}$. Start like you did in the question: you have $7$ choices for $a$, $6$ for $b$ and $5$ for $c$. But remember that $(a \ b \ c) = (b \ c \ a) = (c \ a \ b)$, and hence you must divide the number by $3$. Thus there are $\frac{7 \cdot 6 \cdot 5}{3} = 70$ $3$-cycles.
A: I'm not sure if answering one's own question so close after posting it is a fuax-pas, but I've been struggling with it for days and only posted it here as a last resort for some outside help. I'm leaving my answer here in case others have similar questions that can be solved via the search function.
I realized that there are several ways to write the definition of the centralizer. 
At first I had written it as such, $C_{A_7}((567))=\{\sigma\in A_7|\sigma\tau=\tau\sigma,\forall\tau\in<(567)>\}$, which is the common definition. But after thinking about it further, I realized I could write it in another light, $C_{A_7}((567))=\{\sigma^i\tau|i=1,2,3,\forall\tau\in A_4, \sigma\in <567>\}$. 
Now it becomes trivial as we can define $\Phi(c)=(\sigma^i,\tau)$ where $c\in C_{A_7}((567))$, $\sigma\in<567>$, and $\tau\in A_4$, such that $c=\sigma^i\tau$. Then I was able to show that it was a homomorphism, as well as surjective and injectice. Also, spoiler alert, $\ker \Phi =\{\epsilon\}$, so it defines a nice and neat isomorpishm satisfying the FIT.
