Find all semidirect products of $(\mathbb{Z}_4,+)$ by $C_2$

Problem: Find all semidirect products of $$(\mathbb{Z}_4,+)$$ by $$C_2$$ (the cyclic group of order $$2$$).

My attempt: We know that $$(\mathbb{Z}_4,+)$$ is a cyclic group of order $$4$$. To find all semidirect products of $$(\mathbb{Z}_4,+)$$ by $$\text{C}_2 = \langle {a\rangle}$$ we have to find a homomorphism $$\theta \colon \text{C}_2 \rightarrow \text{Aut}(\mathbb{Z}_4,+)$$. We have $$\text{Aut}(\mathbb{Z}_4,+) \cong \{\sigma_k \mid k \in \mathbb{Z}, (k,4)=1\} = \{\sigma_1,\sigma_3\}$$, $$\text{C}_2 = \{a^0,a^1\} = \{1,a\}$$. So we define $$\theta$$ as follow $$\theta\colon \text{C}_2 \rightarrow \text{Aut}(\mathbb{Z}_4,+)$$, $$\langle a \rangle \mapsto \sigma_1,\sigma_3$$. Group $$G = \{(1,\sigma_1),(1,\sigma_3),(a,\sigma_1),(a,\sigma_3)\}$$ is a semidirect product of $$(\mathbb{Z}_4,+)$$ by $$\text{C}_2$$.

Please check my solution. Is that true? Thank all!

• The order of $G$ should be $8$, not $4$. – Hongyi Huang May 10 at 5:06
• It is either $C_4\times C_2$ or $D_8$. – Hongyi Huang May 10 at 5:07
• But $\text{Aut}(\mathbb{Z}_4,+)$ has four elements. – Minh May 10 at 5:07
• $\mathrm{Aut}(\mathbb{Z}_4)\cong C_2$. – Hongyi Huang May 10 at 5:08
• @HongyiHuang What do you mean? I didn't understand. – Minh May 10 at 5:09

Find $$G = \langle a\rangle:\langle b\rangle\cong\mathbb{Z}_4:\mathbb{Z}_2$$, a semi-direct product.

You correctly find that there is a homomorphism $$\langle b\rangle\to\mathrm{Aut}(\langle a\rangle)\cong\mathbb{Z}_2$$, so $$b^{-1}ab = a^m$$ for $$m = 1$$ or $$m = 3$$.

If $$m = 1$$, then $$a$$ and $$b$$ are commutative. In this case $$G \cong \mathbb{Z}_4\times\mathbb{Z}_2$$.

If $$m = 3$$, then $$bab = a^{-1}$$. This is isomorphic to the dihedral group $$D_8$$.

• A little question: when $m=3$, why $bab=a^{-1}$? – Minh May 10 at 5:49
• $a^3 = a^{-1}$ and $b = b^{-1}$ by definitions. – Hongyi Huang May 10 at 5:49
• From $b^{-1}ab = a^3 \Rightarrow b^2 b^{-1} ab = b^2 a^3 \Rightarrow bab = b^2 a^3 = a^3$, it implies $bab=a^{-1}$ by $a^3 = a^{-1}$, but $\langle a \rangle \cong \mathbb{Z}_4$ has order $4$. – Minh May 10 at 5:58
• Yes, $\langle a\rangle\cong\mathbb{Z}_4$ implies $a^4 = 1$, and so $a^3 = a^{-1}$.@Minh – Hongyi Huang May 10 at 6:00
• Oh! I understand this. – Minh May 10 at 6:01