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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?

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This previous answer discusses how to find a presentation for a semidirect product. In this case, your first semidirect product has presentation $$ \langle a,b,c \mid a^2=b^2 = (ab)^2 = c^7=1,aca^{-1}=c^{-1},bcb^{-1}=c\rangle $$ Here we have used the fact that $c\mapsto c^{-1}$ is the automorphism of $\mathbb{Z}_7$ of order two.

From this presentation, we can see that $b$ commutes with $a$ and $c$, so the group is a direct product of $\langle b\rangle$ and $\langle a,c\rangle$: $$ \langle b \mid b^2 =1 \rangle \times \langle a,c\mid a^2=c^7=1,a^{-1}ca = c^{-1}\rangle \;\cong\; \mathbb{Z}_2\times D_7. $$ This is actually the same as $D_{14}$, since $bc$ has order $14$ and $a^{-1}(bc)a = (bc)^{-1}$. (In general $\mathbb{Z}_2 \times D_n \cong D_{2n}$ when $n$ is odd.)

The other group is given by a similar presentation: $$ \langle x,y \mid x^4 = y^7 = 1,xyx^{-1}=y^{-1}\rangle $$ Note that this is not $D_{14}$, since $D_{14}$ has no elements of order $4$. It is the dicyclic group of order 28.

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  • $\begingroup$ Thanks, its unfortunately been a year since I've done algebra and it was with a far less rigorous book. $\endgroup$
    – Alan
    Jun 22 '15 at 1:01

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