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!

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

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

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