Isomorphism type of order 16 subgroup of $S_6$ [This question is prompted by Robinson ex. 1.6.17 which I am self-studying.
The question is mostly about improving my style and rigour.]
I needed to find the isomorphism type of a Sylow 2 subgroup of $S_6$.
Obviously such a subgroup is of order 16 (the highest power of 2 dividing 6!)
Since all the Sylow 2 subgroups are isomorphic, I just need to pin one down, call it $S$, and investigate it.
First I looked at the representation of $S_6$ as permutations, and spotted that (some) permutations on elements 1..6 elements can be split into a direct sum of permutations on 1..4 and 5..6.  So $S_4 \times S_2$ is a subgroup of $S_6$.
Question: how to make the previous remark more rigorous/elegant?
By Sylow again, $S_4$ has a subgroup of order $8$. So there is a subgroup of order 16 within $S_4 \times S_2$.
Again, looking at $S_4$ as symmetries of 1,2,3,4 and playing around I eventually found elements $a = (1 2)$ and $b = (1 4 2 3)$ with $abab=1$, I realised that $a$ and $b$ generate $D_8$, and so $S \cong D_8\times Z_2$.
Question: how to remove some of the guesswork from this step.  For example, I know $D_8$ is a candidate of the right order, how can I show that it is or isn't the right group more easily than guessing permutations?  [I can usually understand the general theories, but when confronted by specific groups I go to pieces and end up tabulating.]
Thanks.
 A: For the remark that $S_6$ contains a subgroup ismorphic to $S_4\times S_2$, I believe the neatest way of saying it is that the standard embedding of $S_4$ in $S_6$ (i.e. permutations fixing 5 and 6) commutes elementwise with $H=\langle (56)\rangle\simeq S_2$ and their intersection is trivial, so that their product $S_4\cdot H=\{\sigma\tau\ |\ \sigma\in S_4, \tau\in H\}$ is indeed isomorphic to their direct product.
The rest of the argument is fine in itself: moreover, you will have to do some handiwork when dealing with concrete groups. For example, knowing that $D_8$ has the right order is not enough.  Even if you exclude abelian groups of order 8 (which seem unlikely to be 2-Sylows of $S_4$), you are left with two options, namely $D_8$ or $Q_8$ (quaternions), and it is concrete exploration of $S_4$ to give you the answer on which of the two is the correct Sylow group.
Surely, though, there are ways to pick the right permutations that are more clever than others: in this case, moreover, there is a way you can get (somewhat) naturally to the answer, i.e. costruct a $2$-Sylow of $S_n$ inductively on $n$. For a positive integer $n$, call $\mu_2(n)$ the exponent of 2 in the prime factorization of $n!$, $P_n$ the (isomorphism class) of a 2-Sylow of $S_n$.
Clearly, $P_2\simeq C_2$ (where $C_n$ is the cyclic group of order $n$), and this holds for $P_3$ as well since $S_2$ embeds in $S_3$ and $\mu_2(3)=\mu_2(2)$.
As $\mu_2(4)=\mu_2(2)+2$, you must enlarge $\langle(12)\rangle<S_4$ by a factor 4. One factor 2 comes naturally by embedding $S_2$ in $S_4$ as $\langle(34)\rangle$: for the same argument used in the first paragraph, you may clearly see that $$H=\langle(12)\rangle\cdot\langle(34)\rangle\simeq C_2\times C_2.$$ Now, you have no more disjoint $C_2$ to multiply to your $H$, but you can still exchange $(12)$ and $(34)$, namely by conjugating by the double transposition $(13)(24)$. If $K=\langle(13)(24)\rangle$, this gives the product $HK$ (which is a subgroup, since $H$ and $K$ commute) a natural structure of semidirect product $H\rtimes K$ or, if you prefer, the isomorphism structure of $D_8$, so that $P_4\simeq (C_2\times C_2)\rtimes C_2$.
At this point, since $\mu_2(6)=\mu_2(4)+1$, just multiplying $P_4<S_6$ by the disjoint $C_2$ given by $\langle (56)\rangle$ does the trick, and leaves you with $P_6\simeq P_4\times P_2$.
Also, this construction has the advantage of showing a pattern. In fact, if you compute $\mu_2(n)$, you will easily see that:

*

*$P_{2^{n+1}}\simeq (P_{2^{n}}\times P_{2^{n}})\rtimes C_2$ as in the case of $P_4$, which is normally written $P_{2^{n}}\wr C_2$ and called a wreath product;

*if $n=a_0+a_1\cdot 2+\cdots+a_k\cdot 2^k$ in base 2, $P_n\simeq \prod_{j=1}^k (P_{2^i})^{a_i}$, as with $P_6$;

*this generalizes to other primes in the obvious way.

