conceptual question with example: proving groups are isomorphic The question is to prove $D_8$ and the subgroup of $S_4$ generated by $(1 2)$ and $(1 3)(2 4)$ are isomorphic.
I was able to show that the relations for $D_{8}$ follow when we set $b = (1 2)(1 3)(2 4) = (1 3 2 4)$ and $a = (1 2)$ (in particular, $a^2 = b^4 = 1$, $ba = ab^{-1}$). 
I've seen solutions to this problem (and similar problems involving finite groups) that claim isomorphism follows directly from the fulfillment of the above relations, and I've seen much more involved solutions that go into the orders of the subgroups, constructing the actual homomorphism from one group to another (and then claiming it's an isomorphism), and much more.
The variance in solutions has left me confused regarding what exactly I need to show. I would love to get away with a proof as clean as claiming that isomorphism follows directly from the fulfillment of the relations. When can I do so, and why? Also, why does the fact that the relations hold imply that there exists a homomorphism from $D_8$ to this subgroup?
 A: By showing all the relations you actually got that if you have this subgroup of $S_4$ then you can get $D_4$ by simply changing the names of the elements and the name of the operation. In $D_4$ let's say $r$ is the clockwise rotation by angle $\frac{2\pi}{4}$ and $s$ is any reflection. Then $D_4=\langle r,s\rangle$, $r^4=s^2=e,rs=sr^{-1}$. It is clear that if you change the names from $r$ to $b$ and from $s$ to $a$ then you will get the subgroup of $S_4$. So it is clear they are isomorphic.
But if you want a formal solution then you need to really show the isomorphism. First show that every element in the subgroup of $S_4$ can be written in a unique way in the form $b^ia^j$ when $0\leq i\leq 3$, $0\leq j\leq 1$. In the say way every element in $D_4$ can be written in a unique way in the form $r^is^j$ when $0\leq i\leq 3$, $0\leq j\leq 1$. And now you can easily build an isomorphism $\varphi:D_4\to H$ (when $H$ is the subgroup of $S_4$) by $\varphi(r^is^j)=a^ib^j$. Easy to see it is an isomorphism. 
A: By the Sylow theorems we know that $S_4$ has $3$ subgroups of order $8$, namely the $2$-Sylow subgroups, all isomorphic to the dihedral group $D_4$ (also called $D_8$ sometimes) of order $8$. So it is enough to check whether the subgroup generated by $(12)$ and $(13)(24)$ has order $8$.
Reference: How to enumerate subgroups of each order of $S_4$ by hand
Are all Sylow 2-subgroups in $S_4$ isomorphic to $D_4$?
