Find an isomorphism between two groups: $G_1=\langle(123)(45)\rangle$ and $G_2=\{e,(12)(34),(567),(576),(12)(34)(567),(12)(34)(576)\}$ 
Find an isomorphism between two groups: $G_1=\langle(123)(45)\rangle$ and $G_2=\{e,(12)(34),(567),(576),(12)(34)(567),(12)(34)(576)\}$.

I already know the answer to the question, but I have a few questions on how to get there.
First of all, how can I "create" all elements generated by $\langle(123)(45)\rangle$. If it was in $\mathbb Z_n$, I would just add the element mod n until I get back to the identity. But with this, how do I do it?
And then, is there a way to find the isomorphism or you have to try and see? I know the identity will always go to the identity, but other than that?
 A: To answer your first question, the group $\langle (123)(45)\rangle$ is a finite cyclic group, so it very much behaves like the cyclic subgroups of $\mathbb{Z}$, i.e. there is a minimal $k$ such that $((123)(45))^k=e$ after which the elements repeat.
Clearly we have $(123)^3=e$ and $(45)^2=e$, and moreover $(123)^{3m}=e$ and $(45)^{2n}=e$ for any two non-negative integers $m$ and $n$. We are therefore interested in those powers $3m$ and $2n$ for which the cycles $(123)$ and $(45)$ simultaneously go to $e$, i.e. in the minimal solution of $3m=2n$, which is $(m,n)=(2,3)$. In particular, we have
$$((123)(45))^6=(123)^6(45)^6=(123)^{3\cdot 2}(45)^{2\cdot 3}=e$$
and therefore
$$((123)(45))^{6+i}=((123)(45))^i$$
for any non-negative integer $i$.
Therefore to list the non-trivial elements of $\langle (123)(45)\rangle$ we have to figure out
$$((123)(45))^i=(123)^i(45)^i$$
for $i=1,2,3,4,5$, which are
$$(123)(45), (132), (45), (123),\text{ and }(132)(45).$$
A: I am not sure if this is any easier than "seeing it" as you suggest but you see that both groups have order 6. $|G_1| = 6$ because the it is the order of the element $(123)(45)$ which is the least common multiple of the lengths of the cycles (which is 6). For the same reason $(12)(34)(567)$ has order 6 as well and so this element generates $G_2$. Now, if we map $(123)(45)$ to $(12)(34)(567)$ then this mapping extends uniquely to a homomorphism. Since this mapping is bijective we have an isomorphism.
A: In my opinion, the question really hinges on how well you understand cyclic groups.
Remember that the notation $\langle (123)(45) \rangle$ means "the subgroup generated by the permutation $(123)(45)$". Since there is a single element in between the angle brackets, the group $G_1$ is a cyclic group (cyclic groups are those groups generated by a single element).
You are also hopefully aware that there is a unique cyclic group of any given order. So, you know $G_1$ is a cyclic group. If it is to be isomorphic to $G_2$, which we can see has order $6$, then both $G_1$ and $G_2$ must both be the cyclic group of order $6$.
The last conceptual detail is that we can specify any homomorphism from a cyclic group, simply by specifying where one of its generators gets sent. Say we have a cyclic group $C$, a generator $c$ of $C$ (meaning that $\langle c \rangle = C$; "$C$ is generated by $c$"), and a homomorphism $\varphi \colon C \to D$. Then, because each $g \in C$ can be written as some power $c^k$ of $c$, we know that $\varphi(c^k) = \varphi(c)^k$. That is, by choosing $\varphi(c)$, we automatically know what happens to each element of $C$. In particular, can you see why the image $\varphi(C)$ of the whole group $C$ will be cyclic, generated by $\varphi(c)$?

That suggests a strategy: map a generator of $G_1$ to a generator of $G_2$, and we're guaranteed an isomorphism.
With all of that in mind, we need a map $\varphi \colon G_1 \to G_2$. We know that $\varphi(G_1)$ will also be a cyclic group. If we can make sure $\varphi\big((123)(45)\big)$ has order $6$, we know it will generate all of $G_2$ so that $\varphi$ is an isomoprhism (we knew it would be a homomorphism, but this forces it to be a bijection). So, you simply need to find an element of $G_2$ with order $6$; this is where $\varphi$ should send $(123)(45)$ (you have a couple choices).    
You determine the order of elements in a permutation group with the usual "multiplication" (composition) of permutations; e.g.,
$$\big((123)(45)\big)^2 = (123)(45)(123)(45) = (132).$$
