Dеscribе а sеmidirеct prоduсt uр tо isоmоrphism

Question

Dеscribе а sеmidirеct prоduсt uр tо isоmоrphism: $Z_4 ⋊ Z_5$

My solution:

$φ_h(n)=a^hn; a$ - fixed nonzero element $Z_4; h_1,h_2 ∈ Z_5; n_1,n_2 ∈ Z_4$

1. $(n_1, h_1)*(n_2,h_2)=(n_1+h_2,h_1+h_2)$

2. $(n_1, h_1)*(n_2,h_2)=(n_1+(-1)^{h_1}n_2, h_1 + h_2)$

3. $(n_1, h_1)*(n_2,h_2)=(n_1+2^{h_1}n_2, h_1 + h_2)$

4. $(n_1, h_1)*(n_2,h_2)=(n_1+3^{h_1}n_2, h_1 + h_2)$

5. $(n_1, h_1)*(n_2,h_2)=(n_1+4^{h_1}n_2, h_1 + h_2)$

It's right?

Answer 1

Since the semidirect product $\mathbb Z_4\rtimes_\varphi\mathbb Z_{5}$ is uniquely determined by the homomorphism $\varphi: \mathbb Z_5\to\operatorname{Aut}(\mathbb Z_4)\cong\mathbb Z_2$. Note that, $\varphi$ has to be identity map, otherwise $\mathbb Z_2\cong\mathbb Z_5/\ker\varphi$ and $|\ker\varphi|=1$ or $5$, but $|\mathbb Z_5/\ker\varphi|=5$ or $1$, and not equal to $2$ in either way. Contradiction. So the only semidirect product of $\mathbb Z_4\rtimes_\varphi\mathbb Z_5$ is isomorphic to $\mathbb Z_{20}$.