# Understanding semidirect product for group of order 30

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 Answer

(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 $$\mathbb Z_5 \times (\mathbb Z_3 \rtimes_{\tilde\phi_1} \mathbb Z_2)$$. ]

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