I am using Dummit and Foote (pg 181-182) and trying to understand section 5.5 on semidirect product for a group of order 30. They start out with a reminder that if $H$ is of order 15 it is a normal subgroup (because of index 2) as well as cyclic, thus we can form a group $G=HK$ where $H\cap K = 1$ then $G\cong H\rtimes K$ for some $\phi:K\rightarrow Aut(Z_{15})$ where $Aut(Z_{15})\cong(\mathbb Z/15 \mathbb Z)^{\times}\cong Z_4\times Z_2$ (I understand how they got this).

Then the book explains that the given $Aut(H)$ has three elements of order 2, one of those elements whose actions on $H = \langle a\rangle\times \langle b\rangle$ is $\{a\mapsto a, b\mapsto b^{-1}\}$ (there are two others the book shows but I just want to understand one of first). If a given $K=\langle k\rangle$ with the action illustrated above the book says that $G = H\rtimes_{\phi_1}K\cong Z_5\times D_6$. My questions are the following:

1) I am not completely sure why $Aut(H)$ has three elements of order $2$ precisely.

2) I do not understand the final conclusion, where did the dihedral group come from and where did $Z_5$ come from?

I think if I understand this, the other two non-isomorphic groups should make more sense. Thanks!


(1) For an element $(m, n) \in \mathbb Z_4 \times \mathbb Z_2$ to be of order two, it must be the case that $(2m, 2n) = (0, 0)$ in $\mathbb Z_4 \times \mathbb Z_2$, i.e. $m \in \{0, 2 \}$ and $n \in \{ 0, 1 \}$. The case $(m, n) = (0,0)$ is actually of order one; the three remaining cases however are genuinely of order two. The case that you are considering is $(m, n) = (0, 1)$ (and to see this, remember that the isomorphism between $\mathbb Z_2$ and $U_3$ (the multiplicative group of units modulo $3$) sends $0 \mapsto 1$ and $1 \mapsto -1$).

(2) You know that $H = \mathbb Z_{15} = \mathbb Z_{5} \times \mathbb Z_3$, and you have described an action $\phi_1$ of $K = \mathbb Z_2$ on $ \mathbb Z_5 \times \mathbb Z_3 $, in which the $\mathbb Z_2$ acts trivially on the $\mathbb Z_5$ factor, but acts non-trivially on the $\mathbb Z_3$ factor (with the non-trivial element in $\mathbb Z_2$ sending the generator $b$ of $\mathbb Z_3$ to $b^{-1}$). Hence the semidirect product $H \rtimes_{\phi_1} K$ is isomorphic to $\mathbb Z_5 \times (\mathbb Z_3 \rtimes_{\tilde\phi_1} \mathbb Z_2)$, where $\tilde\phi_1$ is the action of $\mathbb Z_2$ on $\mathbb Z_3$ in which the non-trivial element in $\mathbb Z_2$ sends the generator $b$ of $\mathbb Z_3$ to $b^{-1}$. But this $\mathbb Z_3 \rtimes_{\tilde\phi_1} \mathbb Z_2$ is precisely $D_6$, as required.

[Let me spell out the crucial step in the argument in a bit more detail. Consider the multiplication rule in $(\mathbb Z_5 \times \mathbb Z_3) \rtimes_{\phi_1} \mathbb Z_2$. This says that $$ ((p_1, q_1), k_1) . ((p_2, q_2), k_2) = \left((p_1, q_1) . (\phi_1)_{k_1}(p_2, q_2) \ , \ k_1 . k_2\right)$$ where $p_1, p_2 \in \mathbb Z_5$, $q_1, q_2 \in \mathbb Z_3$ and $k_1, k_2 \in \mathbb Z_2$.

But $$(\phi_1)_{k_1} (p_2, q_2) = (p_2, (\tilde \phi_1)_{k_1} (q_2)), $$ and this means that $$((p_1, q_1), k_1) . ((p_2, q_2), k_2) = \left( (p_1 . p_2 \ , \ q_1 . (\tilde \phi_1)_{k_1} (q_2)) \ , \ k_1 . k_2\right),$$ which is precisely the multiplication rule for $\mathbb Z_5 \times (\mathbb Z_3 \rtimes_{\tilde\phi_1} \mathbb Z_2)$. ]

  • $\begingroup$ Thank you so much for the detailed response! I think this is exactly what I needed to understand this! $\endgroup$
    – InsigMath
    Nov 11 '18 at 20:29
  • $\begingroup$ @InsigMath No problem! Good luck with the other two cases! $\endgroup$
    – Kenny Wong
    Nov 11 '18 at 20:30

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