## Classification of groups of order 20

Classify the groups of order 20. [Hint: There are five isomorphism types.]

Note that $20 = 2^2 \cdot 5$.

By FTFGAG, there are two distinct abelian groups of order 20: $Z_{20}$ and $Z_{10} \times Z_2$.

Now let $G$ be a nonabelian group of order 20. By Sylow’s Theorem, $n_5 = 1$, so that $G$ has a unique (hence normal) Sylow 5-subgroup $H \cong Z_5$. Now let $K$ be any Sylow 2-subgroup of $G$. By Lagrange, we have $H \cap K = 1$, so that $G = HK$. By the recognition theorem for semidirect products, $G \cong H \rtimes_\varphi K$ for some $\varphi : K \rightarrow \mathsf{Aut}(H)$. Evidently, classifying the nonabelian groups of order 20 is equivalent to determining the nonisomorphic groups constructed in this manner. To that end, let $H = Z_5 = \langle y \rangle$. Note that $\mathsf{Aut}(H) = \langle \alpha \rangle \cong Z_4$; where $\alpha(y) = y^2$.

Let $K = Z_4 = \langle x \rangle$. There are four distinct homomorphisms $K \rightarrow \mathsf{Aut}(H)$.

If $\varphi_1(x) = 1$, then $\varphi_1$ is trivial; this contradicts the nonabelianicity of $G$.

If $\varphi_2(x) = \alpha$, then $Z_5 \rtimes_{\varphi_2} Z_4$ is indeed a nonabelian group of order 20.

If $\varphi_3(x) = \alpha^2$, then $Z_5 \rtimes_{\varphi_3} Z_4$ is indeed a nonabelian group of order 20. Moreover, since $\mathsf{ker}\ \varphi_3 \cong Z_2$ and $\mathsf{ker}\ \varphi_2 \cong 1$, $H \rtimes_{\varphi_3} K \not\cong H \rtimes_{\varphi_2} K$.

If $\varphi_4(x) = \alpha^3$, then $\mathsf{im}\ \varphi_4 = \mathsf{im}\ \varphi_2$. Since $Z_4$ is cyclic, by a previous theorem, $H \rtimes_{\varphi_4} K \cong H \rtimes_{\varphi_2} K$.

Thus there are two distinct groups of order 20 which have a cyclic Sylow 2-subgroup.

Suppose now that $K = Z_2^2 = \langle a \rangle \times \langle b \rangle$. Again, $\psi : Z_2^2 \rightarrow Z_4$ is determined uniquely by $\psi(a)$ and $\psi(b)$, and is indeed a homomorphism provided $|\psi(a)|$ and $|\psi(b)|$ divide 2. We thus have $\psi(a), \psi(b) \in \{ 1, \alpha^2 \}$, for a total of four choices.

If $\psi_1(a) = \psi_1(b) = 1$, then $\psi_1 = 1$, contradicting the nonabelianicity of $G$.

If $\psi_2(a) = \alpha^2$ and $\psi_2(b) = 1$, then $Z_5 \rtimes_{\psi_2} Z_2^2$ is indeed a nonabelian group of order 20.

If $\psi_3(a) = 1$ and $\psi_3(b) = \alpha^2$, then $\varphi_3 = \varphi_2 \circ \theta$, where $\theta(a) = b$ and $\theta(b) = a$. Clearly $\theta$ is an automorphism of $Z_2^2$. By a lemma to a previous theorem, we have $H \rtimes_{\psi_3} K \cong H \rtimes_{\psi_2} K$.

If $\psi_4(a) = \alpha^2$ and $\psi_4(b) = \alpha^2$, then $\varphi_4 = \varphi_2 \circ \theta$, where $\theta(a) = a$ and $\theta(b) = ab$. Clearly $\theta$ is an automorphism of $Z_2^2$. By a lemma to a previous theorem, we have $H \rtimes_{\psi_4} K \cong H \rtimes_{\psi_2} K$.

Thus there is a unique nonabelian group of order 20 which has an elementary abelian Sylow 2-subgroup.

In summary, the distinct groups of order 20 are as follows. We let $Z_5 = \langle y \rangle$, $Z_4 = \langle x \rangle$, and $Z_2^2 = \langle a \rangle \times \langle b \rangle$.

1. $Z_{20}$
2. $Z_{10} \times Z_2$
3. $Z_5 \rtimes_{\varphi_3} Z_4$, where $\varphi_3(x)(y) = y^{-1}$.
4. $Z_5 \rtimes_{\varphi_2} Z_4$, where $\varphi_2(x)(y) = y^2$
5. $Z_5 \rtimes_\psi Z_2^2$, where $\psi(a)(y) = y^{-1}$ and $\psi(b)(y) = y$.