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Can I write elements of $G=C_3\rtimes C_2$ as $$\{(0,0),(0,1),(1,0),(1,1),(2,0),(2,1)\}?$$

Then, what are the generators of $G$? $(0,1)$ and $(1,0)$?

I've learned that the multiplication of semi-direct product is defined as $$(n_1,h_1)(n_2,h_2)=(n_1(h_1n_2h_1^{-1}),h_1h_2),$$

but since $C_3$ and $C_2$ are both abelian, it just becomes $$(n_1,h_1)(n_2,h_2)=(n_1n_2,h_1h_2).$$

This makes $G$ to be a direct product, not a semi-direct product. What is the correct way to define a multiplication on $G$?

Shaun
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user642721
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1 Answers1

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(1) Let $K$ be a group, let $H$ be a group, and let $\phi:H\to\text{Aut}(K)$ be a group homomorphism. Here $\text{Aut}(K)$ is the group of automorphisms of $K$. For each $h\in H$, $\phi$ will send $h\mapsto\phi_h$. The semidirect product $K\underset{\phi}{\rtimes}H$ is then defined to be the group whose underlying set is $K\times H$, and whose multiplication is defined as follows: for every $(k_1,h_1),(k_2,h_2)\in K\underset{\phi}{\rtimes}H$,

$$(k_1,h_1)\cdot(k_2,h_2)=(k_1\cdot\phi_{h_1}(k_2),h_1\cdot h_2).$$

(2) Note that for any groups $H$ and $K$, there is always at least one homomorphism $H\to\text{Aut}(K)$, namely the map $\phi_0:H\to\text{Aut}(K)$ which sends each element of $H$ to the trivial automorphism. You can check that for any groups $H$ and $K$, we have that $K\underset{\phi_0}{\rtimes}H$ is equal to the direct product $K\times H$.

(3) Let $K=C_3$ and $H=C_2$. Note that $\text{Aut}(C_3)$ consists of two functions: the trivial automorphism $1_{C_3}$, and the map $f$ which sends $x\mapsto2x$ for each $x\in C_3$. It follows that there are exactly two homomorphisms $C_2\to\text{Aut}(C_3)$: there's the trivial one which sends each $x\mapsto1_{C_3}$ for all $x\in C_2$, and there's the other one which sends $0\mapsto1_{C_3}$ and $1\mapsto f$. Since there are exactly two homomorphism $C_2\to\text{Aut}(C_3)$, one of which is trivial, when people write $C_3{\rtimes}C_2$, they always mean $C_3\underset{\phi}{\rtimes}C_2$, where $\phi:C_2\to\text{Aut}(C_3)$ is the non-trivial homomorphism.

(4) We can say more. In fact $C_3{\rtimes}C_2=\{(a,b)\,\vert\,a\in C_3,b\in C_2\}$, and for all $(a_1,b_1),(a_2,b_2)\in C_3\rtimes C_2$, we have that

$$(a_1,b_1)\cdot(a_2,b_2)=(a_1+2^{b_1}\cdot a_2,b_1+b_2).$$

user729424
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