No element of $\mathbb Z_{10}$ has order four (why not?) and there is one element of order 2 ($5\in\mathbb Z_{10}$ under addition, $10\cong -1\in \mathbb Z_{11}^x$ under multiplication.), so our possibilities are quite limited. There are two groups of order $4$, and either will work as our $R$. To give a nonabelian group, we need to pick a nontrivial homomorphism, as you pointed out.
So at least one generator of $R$ has to map to our order-two element. In the case of the Klein four group, there appear to be three possibilities, but I claim that up to an isomorphism of the Klein four group, there is only one possibility.
Actually, this exhausts the possibilities for groups of order 44: we have two abelian groups, $\mathbb Z_{11}\times\mathbb{Z}_4$, $\mathbb Z_{11}\times \mathbb{Z}_2\times \mathbb Z_2$, and two nonabelian groups: $\mathbb{Z}_{11} \rtimes \mathbb Z_4 = \langle a,b \mid a^{11}, b^4, b^{-1}ab = a^{-1}\rangle$ and $\mathbb Z_{11}\rtimes(\mathbb Z_2 \times \mathbb Z_2) = \langle a, b, c \mid a^{11}, b^2, c^2, [b,c], b^{-1}ab = c^{-1}ac = a^{-1} \rangle$. I think the latter is $D_{22}$ (for 22-gon, not order of group), while the former has an element of order $4$.
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.
Best Answer
Let's assume that the number of 3-Sylow subgroups is 4*
Let $G$ act on the set of 3-Sylow subgroups, and this will give a homomorphism $$ f : G\to S_4 $$ We claim that this homomorphism is injective, and so an isomorphism.
a) Let $K = \ker(f)$, then $K < N_G(P)$ where $P$ is some fixed 3-Sylow subgroup. Now, $$ [G:N_G(P)] = 4 \Rightarrow |N_G(P)| = 6 \Rightarrow N_G(P) \cong \mathbb{Z}_6 \text{ or } S_3 $$ If $N_G(P) \cong \mathbb{Z}_6$, then $G$ would have an element of order 6, and so $N_G(P) \cong S_3$, and so $$ |K| \in \{1,3,6\} $$ (Note $|K| \neq 2$ since $S_3$ has no normal subgroups of order 2)
b) If $|K| = 1$ we're done and if $|K| = 6$, then $K = N_G(P) \triangleleft G$, but $N_G(N_G(P)) = N_G(P)$ which is a contradiction. Hence, assume $|K| = 3$
c) If $|K| =3$, then by the N/C theorem, $$ G/C_G(K) = N_G(K)/C_G(K) \cong \text{ a subgroup of } Aut(\mathbb{Z}_3) \cong \mathbb{Z}_2 $$ In particular, $2 \mid C_G(K)$, so $C_G(K)$ contains an element of order 2, which, when multiplied with a generator of $K$ will give an element of order 6 - a contradiction.
Thus, $|K| = 1$ and we're done.
$\ast$ All that remains to show is that $n_3 = 4$, or equivalently, $n_3 \neq 1$ : I believe that a unique (and hence normal) 3-Sylow subgroup would end up producing an element of order 6 (perhaps by looking at the product $HK$ for some suitable $K$), but I am not sure about that.