$C_4\rtimes C_2$ isn't isomorphic to $C_4\times C_2$

Question

Explain why $C_4\rtimes C_2$ isn't isomorphic to $C_4\times C_2$

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My first thought was to say that since $C_4\rtimes C_2 \simeq C_4$ then if $C_4\rtimes C_2\simeq C_4\times C_2 $ we would get $C_4\times C_2 \simeq C_4$ which is a contradiction since $|C_4\times C_2|\neq |C_4|$

Is my answer correct ?

Is there a better way/answer to explain this ?

(I am assuming the exercise means $C_4,C_2$ are the cyclic groups with $4 $ and $2$ elements)

Answer 1

One is abelian, the other is not (assuming the action defining the semidirect product is not trivial).

Theorem: If $\varphi: G\to H$ is a surjective group homomorphism and $G$ is abelian, then $H$ is abelian.

Proof: Let $x,y\in H$. Then there exist $a,b\in G$ such that $x=\varphi(a), y=\varphi(b)$, so

$$\begin{align} xy&=\varphi(a)\varphi(b)\\ &=\varphi(ab)\\ &=\varphi(ba)\\ &=\varphi(b)\varphi(a)\\ &=yx. \end{align}$$

Hence $H$ is abelian.$\square$