I have been thinking about a composition series for $D_{14}\times D_{10}$ (where $D_{2n}$ is the dihedral group with $2n$ elements). Is the following a correct composition series for $D_{10}\times D_{14}$: $$D_{14}\times D_{10}\vartriangleright \langle\sigma_1\rangle\times D_{10}\vartriangleright\langle\sigma_1\rangle\times\langle\sigma_2\rangle\vartriangleright\{id_1\}\times\{id_2\}?$$ I also have to verify what the factors are, and in this case, I think the factors are $$\begin{align}&(D_{14}\times D_{10})/(\sigma_1\times D_{10})\cong\mathbb{Z}_2\\& (\langle\sigma_1\rangle\times D_{10})/(\langle\sigma_1\rangle\times\langle\sigma_2\rangle)\cong\mathbb{Z}_2\\&\langle\sigma_1\rangle\times\langle\sigma_2\rangle\cong\mathbb{Z}_{35}.\end{align}$$ I'm really not sure I understand the material very well, so if possible, please try to elaborate as much as you can. Thanks!

Please tell me if this is correct, and if it isn't, what I should change (and if maybe it isn't so clear, why it should be changed).

  • $\begingroup$ The quotients must also be simple. But $\mathbb Z/\mathbb Z{35}$ is not simple. $\endgroup$ – Hans Giebenrath Dec 15 '12 at 19:48
  • $\begingroup$ Ah! So I should insert $\langle\sigma_1\rangle\times\{id_2\}$ between the final two, right? I knew that, for some reason I just didn't think about it. Everything else seem okay? $\endgroup$ – Clayton Dec 15 '12 at 19:52
  • $\begingroup$ I think this should do it. I think you should also write what $\sigma_i$ actually is. I'm not sure how canonical the notation for dihedral groups is. $\endgroup$ – Hans Giebenrath Dec 15 '12 at 19:56
  • $\begingroup$ It has been the standard notation in my course/textbook. I do apologize, but I think for the upcoming exam, such a thing would be unambiguous. Thanks for reviewing it. If you post your thoughts/comments, I'll accept it. $\endgroup$ – Clayton Dec 15 '12 at 20:02
  • $\begingroup$ If $\sigma_1$ denotes "generator" of order $n$, then the quotient $D_{14} / \langle \sigma_1 \rangle$ should be isomorphic to $\mathbb Z/\mathbb Z 2$. If it is the generator of order $2$, then the subgroup $\langle \sigma_1 \rangle$ is not normal. $\endgroup$ – Hans Giebenrath Dec 15 '12 at 20:13

Your composition series looks valid except for the last quotient. You should insert a $\langle \sigma_1 \rangle \times \{ 1 \}$ (or $\{ 1 \} \times \langle \sigma_2 \rangle$). Then you will get two additional quotients isomorphic to the simple groups $\mathbb Z / 5\mathbb Z$ and $\mathbb Z / 7\mathbb Z$.


Composition series for direct products

I think there is a more systematic answer to your question. Assume you have two groups $G$ and $H$ with composition series $$ G = G_r \supsetneq G_{r-1} \supsetneq \dotsb \supsetneq G_0 = \{ 1 \} $$ and $$ H = H_s \supsetneq H_{s-1} \supsetneq \dotsb \supsetneq H_0 = \{ 1 \}.$$ Then we obtain a chain $$ G \times H = G_r \times H \supsetneq \dotsb \supsetneq \{ 1\} \times H = \{ 1 \} \times H_s \supsetneq \{ 1 \} \times H_{s-1} \supsetneq \dotsb \supsetneq \{ 1 \} \times \{ 1 \}. $$ It is easy to see that this is a composition series of $G \times H$. In particular we conclude that the composition length of $G \times H$ is the sum of the composition length of $G$ and $H$.

How does this apply in our situation? We have two composition series $$ D_{14} \supsetneq \langle \sigma_1 \rangle \supsetneq \{ 1 \} $$ and $$ D_{10} \supsetneq \langle \sigma_2 \rangle \supsetneq \{ 1 \}. $$ Thus we obtain the following composition series for $D_{14} \times D_{10}$: $$ D_{14} \times D_{10} \supsetneq \langle \sigma_1 \rangle \times D_{10} \supsetneq \{ 1 \} \times D_{10} \supsetneq \{ 1 \} \times \langle \sigma_2 \rangle \supsetneq \{ 1 \} \times \{ 1 \}. $$

I think the same can be done for semidirect products $N \rtimes H$ if one knows composition series for $N$ and $H$.

  • $\begingroup$ @Clayton: I think now it's a more useful answer. $\endgroup$ – Hans Giebenrath Dec 16 '12 at 6:55

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