So, I'm very new to working with semi-direct products. I'm working on my algebra qual prep, and one of the questions was to identify all the groups of order 28. I'm pretty sure I have the answer...the only problem is I'm not sure just what two of the groups look like!
So, starting with the sylow subgroups, I have that $n_2|7$ and $2|(n_2-1)$, so we have $n_2=1$ or $n_2=7$. Similarly, $n_7|4$ and $7|(n_7-1)$, so $n_7=1$, thus the Sylow 7 subgroup is unique, and therefore normal, hence our group G is a semidirect product of its Sylow 2 subgroup and its sylow 7 subgroup.
The two possibilities for the Sylow 2 subgroup are $\mathbb Z _2 \times \mathbb Z _2$ or $\mathbb Z_4$. Neither are eliminated by a counting argument. So, I'm looking for homomorphisms from the sylow 2 subgroup to the automorphism group of the sylow 7 subgroup, which is isomorphic to $\mathbb Z_6$
Case 1: The sylow 2 subgroup is $\mathbb Z_2 \times \mathbb Z_2$. The generators are $(0,1)$ and $(1,0)$. These each have order 2, so the images must have order $1$ or $2$. So the images are either $0$ or $3$. If the images of both are $0$, we have the trivial homomorphism, so this is actually a direct product and we have the abelian group $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_7$ Now, say $(1,0)$ goes to $3$ and $(0,1)$ goes to $0$. Then $(1,1)$ goes to 3. Similarly, we have as a case $(0,1)\to 3,(1,0)\to 0,(1,1)\to 3$ and $(1,0)\to 3,(0,1)\to 3,(1,1)\to 0$. All of these are identical up to isomorphism as they send 2 indistinguishable elements to the same thing and the other to the identity, so up to isomorphism there is only one such semidirect product...
Here's where I'm stuck, I don't know what this group is, what it looks like! How do I figure that out?
Case 2: The sylow 2 subgroup is $\mathbb Z_4$. Here we just have one generator, $1$, and it can go to either $0$ or $3$ as above. If $0$ we have the abelian case. What does the nonabelian case look like?