# Recognizing semidirect products in classification of groups of a given order

I am reading about classifications of groups of order $$30$$ and $$12$$ in Dummit and Foote. I understand the general procedure for such classifications. However, I have trouble recognizing the resulting semidirect products as "its more descriptive form"; for example, as a direct product of a cyclic group and a dihedral group. Below are the specific examples:

1. Let $$G$$ be a group of order $$30$$. Let $$H = \langle a\rangle \times \langle b \rangle \cong Z_5 \times Z_3$$ be normal subgroup of $$G$$ of order $$15$$. Let $$K = \langle k \rangle$$ be the Sylow $$2$$- subgroup of $$G$$.

Let $$\phi_1: K \to \mbox{Aut}(H)$$ that maps $$k$$ to $$\{a \mapsto a, b \mapsto b^{-1}\}$$. Then $$G_1 = H\rtimes_{\phi_1}K \cong Z_5 \times D_6$$ (note that in this semidirect product $$k$$ centralizes the element $$a$$ of $$H$$ of order $$5$$, so the factorization as a direct product is $$\langle a \rangle \times \langle b,k \rangle$$).

Let $$\phi_2: K \to \mbox{Aut}(H)$$ that maps $$k$$ to $$\{a \mapsto a^{-1}, b \mapsto b\}$$. Then $$G_2 = H\rtimes_{\phi_2}K \cong Z_3 \times D_{10}$$ (note that in this semidirect product $$k$$ centralizes the element $$b$$ of $$H$$ of order $$3$$, so the factorization as a direct product is $$\langle b \rangle \times \langle a,k \rangle$$).

Let $$\phi_3: K \to \mbox{Aut}(H)$$ that maps $$k$$ to $$\{a \mapsto a^{-1}, b \mapsto b^{-1}\}$$. Then $$G_3 = H\rtimes_{\phi_2}K \cong D_{30}$$.

1. Let $$G$$ be a group of order $$12$$. Let $$V$$ be a Sylow $$2$$-subgroup of $$G$$ and $$T$$ be a Sylow $$3$$-subgroup of $$G$$. Suppose $$T \trianglelefteq G$$ and $$V = \langle a\rangle \times \langle b \rangle \cong Z_2 \times Z_2$$. Put $$\mbox{Aut}(T) = \langle \lambda \rangle \cong Z_2$$. Then there are three nontrivial homomorphisms from $$V$$ into $$\mbox{Aut}(T)$$ determined by specifying their kernels as one of the three subgroups of order $$2$$ in $$V$$. For example, $$\phi_1(a) = \lambda$$ and $$\phi_1(b) = \lambda$$ has kernel $$\langle ab \rangle$$. If $$\phi_2$$ and $$\phi_3$$ have kernels $$\langle a \rangle$$ and $$\langle b \rangle$$, respectively, then the resulting three semidirect products are all isomorphic to $$S_3 \times Z_2$$, where the $$Z_2$$ direct factor is the kernel of $$\phi_i$$. For example, $$T \rtimes_{\phi_1} V = \langle a, T \rangle \times \langle ab \rangle$$.

Can someone offer a more detailed explanation for the bold parts? Are there any results that I am not aware of that make the conclusion follow easily?

$$H$$ is a subgroup of $$G$$ which is the direct product of $$C_5$$ and $$C_3$$. Both $$C_5$$ and $$C_3$$ have an automorphism of order $$2$$, which acts as $$\phi(x)=x^{-1}$$.

Note that $$G/N$$ has order $$2$$, so either $$C_G(N)=N$$ or $$C_G(N)=G$$. In the latter case, you get a direct product $$C_5 \times C_3 \times C_2 = C_{30}$$. So assume that $$C_G(N)=N$$. Now you can define a map $$\phi: C_2 \to \operatorname{Out}(H)$$ as the map defined by the action of $$G/N$$ by conjugation on $$N$$ (this is the NC theorem). There are three possibilities for $$G/N = C_2$$:

• It acts trivially on $$C_5$$, but not on $$C_3$$. In this case you get $$C_5 \times (C_3 \rtimes C_2) = C_5 \times D_6$$.
• It acts trivially on $$C_3$$, but not on $$C_5$$. In this case you get $$C_3 \times (C_5 \rtimes C_2) = C_5 \times D_{10}$$.
• It acts non-trivially on both $$C_5$$ and $$C_3$$. In this case, you get $$(C_5 \times C_3) \rtimes C_2 = C_{15}\rtimes C_2 = D_{30}$$.

Essentially, you are looking at how a $$C_2$$ can act on two factors, each of which has a single automorphism of order two (so the action is either that automorphism, or trivial).

For the second one, it's the same logic, but now you are looking at maps $$\phi: C_2 \times C_2 \to C_2$$ (where the latter is $$\operatorname{Aut}(C_3)$$). Let $$C_2 \times C_2 = \langle a \rangle \times \langle b \rangle$$. Take the preimage of $$C_2 = \operatorname{Aut}(C_3)$$: it can be one of three subgroups $$\langle a \rangle$$, $$\langle b \rangle$$, $$\langle ab \rangle$$. Call it $$T$$. Whatever it is, you can still write:

$$G = (C_3 \rtimes T) \times C_2$$

as follows

• $$G = (C_3 \rtimes \langle a \rangle) \times \langle b \rangle$$
• $$G = (C_3 \rtimes \langle b \rangle) \times \langle a \rangle$$
• $$G = (C_3 \rtimes \langle ab \rangle) \times \langle a \rangle$$

and so all the semidirect products are isomorphic.

• As a tl;dr, if $G=H \rtimes_\phi T$, then the direct factors of $H$ on which $\phi$ acts trivially remain direct factors of $G$. Jun 27, 2021 at 10:56
• By $N$, do you mean $H$, the subgroup of order 15? Can you explain more about the first two sentences of the second paragraph? Why does index=2 imply the two possibilities about the centralizer of H? And why does $C_G(H) = G$ imply $G$ is abelian? Jun 27, 2021 at 11:37
• In addition, for the second one, the book seems to be saying $G = (C_3 \rtimes C_2) \times K$, where $C_2 = \langle a \rangle$ or $\langle b \rangle$ and $K$ is the kernel of $\phi$. This looks slightly different from yours. Are the two isomorphic? Jun 27, 2021 at 11:42
• @mashedcarrots Because since $H$ is abelian $H \leq C_G(H)$, and $[G:H]=2$. So it either $H$ or $G$, there are no intermediate subgroups. It does not immediately imply that $G$ is abelian. However, it does tell you that the semidirect product is actually direct... and hence that $G$ is abelian. Jun 27, 2021 at 12:08
• @mashedcarrots the direct factor acts trivially on $C_3$, and hence it is in the kernel of the map $\phi: C_2 \times C_2 \to \operatorname{Aut}(C_2)$. So yes, they are isomorphic (they are in fact the same group). Jun 27, 2021 at 12:10