Looking for help to understand group isomorphisms I recently posted a question asking about subgroups of symmetric group, and it seemed that all responses went way beyond me and when I attempted to understand, it just seemed like there was a conversation going on that I could not understand. So I decided maybe this is partially on my part, as I should provide more information of what I understand etc. I will also cut it down.
In a few words; I am wondering about the relation between isomorphism and subgroups, how do they relate? How can I use it to help with the following, etc
$H=\{1,(423),(432),(42),(43),(23)\}$
Specifically, I want to prove or disprove that
H Is a subgroup of $S_4$
I know that to test this, it must be that it contains the identity , which it does.
It also must be that if we multiply any two elements, their product is also in H and finally that each element has an inverse in H.
It seems like a lot to test each thing;
Here is what else was suggested at, but not elaborated so I do not know if it is what was meant.
What if I say define a map as follows;
$f: H \to H'$
$(1) \to (1) , (423) \to (123) , (432) \to (132) , (42) \to (12) , (43) \to (13) , (23) \to (23)$
Then it is bijective map. And from a theorem in my notes, it follows that the groups of permutations ,  are isomorphic . But I do not know what this can tell me.
Other things I know about :
Lagranges thereom, that the order of the subgroup must divide the order of the group.
Cayleys thereom, that if G is a finite group of order n then G is isomorphic to $S_{n}$
 A: It seems like you're going down a reasonable path, but haven't done it in quite the right way. In particular, the principal observation is that the subgroup
$$H=\{1,(423),(432),(42),(43),(23)\}$$
is exactly the permutations on the set $\{2,3,4\}$. This is, of course, closed under composition since the composition of two permutations on a set is a permutation on the same set - and we have inverses as well by definition of permutation, as well as an identity element. You can equivalently think of this as the set of permutations which have $1$ as a fixed point.
Defining a bijection from $H$ to $S_3$ is a good idea, but the key point is that this is an isomorphism - that is, it is a bijection satisfying $f(ab)=f(a)f(b)$. You don't really save yourself from having to check that $f(ab)=f(a)f(b)$ for all pairs $a$ and $b$ by defining it pointwise rather than according to a pattern. However, when you see that $f$ just replaces all '4's with '1's, it's clear that it preserves the group operation (i.e. is a homomorphism) and with bijectivity it's clera that it's an isomorphism - at which point, you can see that $H$ is a group since it is isomorphic to a group.
A: There is not so much to check to determine that $H$ is a subgroup of $S_4$.For example the inverses of 2-cycles are themselves and for the 3-cycles it's easy to see that they are each other inverses, and for the calculation of the product there is a similar reasoning.
But there are other solutions: 
-If you want to work with isomorphisms you a priori don't know that $H$ is a group and so we cannot apply the usual rules of isomorphism. You know only that there is a bijection between $H$ and $H'$ which is actually $S_3$. But we can notice that the number $4$ in $H$ has the same, identic function of the element $1$ in $H'=S_3$, like they are mute variables, only the symbol has changed, so the multiplication in $H$ works the same as the multiplication in $S_3$ and so for the structure of $H$ it is a subgroup of $S_4$ in particular $H$ is isomorphic to $S_3$. 
-Another way, that is quite the same of the above solution, is to see at the problem from another point of view. We can look at $H$ as the set of permutation of ${2,3,4}$, that has $3!=6$ elements, and is a group because every permutation group is a group.
A: The map you mention defines $H$ as in bijection with $S_3$, and this map is compatible with the composition of parmutations. Indeed, $H$ consists of all permutations of $\{2,3,4\}$. It is a group on its own right for the composition of permutations, and the canonical injection of $H$ in $S_n$ for any $n\ge 4$ is a group homomorphism. Hence $H$ is isomorphic to a subgroup of $S_n$ for any $n\ge 4$, and even is a subgroups if $n=4$.
