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?


1 Answer 1


$H$ 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.

  • $\begingroup$ 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 $G$. $\endgroup$ Jun 27, 2021 at 10:56
  • $\begingroup$ 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? $\endgroup$ Jun 27, 2021 at 11:37
  • $\begingroup$ 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? $\endgroup$ Jun 27, 2021 at 11:42
  • 1
    $\begingroup$ @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. $\endgroup$ Jun 27, 2021 at 12:08
  • 1
    $\begingroup$ @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). $\endgroup$ Jun 27, 2021 at 12:10

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