Find four groups of order 20 not isomorphic to each other.

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

By Fundamental theorem of Finitely Generated Abelian Group, there are two distinct abelian groups of order $20$: $\mathbb{Z}_{20}$ and $\mathbb{Z}_{10} \times \mathbb{Z}_2$.

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 $H \cong \mathbb{Z}_5$. 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 $H = \mathbb{Z}_5 = \langle y \rangle$. Note that $\mathsf{Aut}(H) = \langle \alpha \rangle \cong \mathbb{Z}_4$; where $\alpha(y) = y^2$.

Let $K = \mathbb{Z}_4 = \langle x \rangle$. 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 $\mathbb{Z}_5 \rtimes_{\varphi_2} \mathbb{Z}_4 $is indeed a nonabelian group of order $20$.

If $\varphi_3(x) = \alpha^2$, then $\mathbb{Z}_5 \rtimes_{\varphi_3} \mathbb{Z}_4$ is indeed a nonabelian group of order $20$. Moreover, since $\mathsf{ker}\ \varphi_3 \cong \mathbb{Z}_2$ 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 $\mathbb{Z}_4$ 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 $K = \mathbb{Z}_2^2 = \langle a \rangle \times \langle b \rangle$. Again, $\psi : \mathbb{Z}_2^2 \rightarrow \mathbb{Z}_4$ 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 $\mathbb{Z}_5 \rtimes_{\psi_2} \mathbb{Z}_2^2$ 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 $\mathbb{Z}_2^2$. 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 $\psi_4 = \psi_2 \circ \theta$, where $\theta(a) = a$ and $\theta(b) = ab$. Clearly $\theta$ is an automorphism of $\mathbb{Z}_2^2$. 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 $\mathbb{Z}_5 = \langle y \rangle$, $\mathbb{Z}_4 = \langle x \rangle$, and $\mathbb{Z}_2^2 = \langle a \rangle \times \langle b \rangle$.

$Z_{20}$,

$Z_{10} \times Z_2$,

$Z_5 \rtimes_{\varphi_3} Z_4$, where $\varphi_3(x)(y) = y^{-1}$.

$Z_5 \rtimes_{\varphi_2} Z_4$, where $\varphi_2(x)(y) = y^2$

$Z_5 \rtimes_\psi Z_2^2$, where $\psi(a)(y) = y^{-1}$ and $\psi(b)(y) = y$.

(Source: Crazyproject)

In addition to the two cyclic groups and the dihedral group $D_{10}$, consider the group $G$ of symmetries of an antiprism over the regular pentagon. It is transitive on the 10 vertices and the stabilizer of a vertex $p$ contains a nontrivial symmetry given by the reflection in the vertical plane passing through $p$ and the origin; for a total of 20 elements. This is different from $D_{10}$ because the central element (reflection in the origin in $\mathbb{R}^3$) has a 5th root in $G$, which is not the case for $D_{10}=D_5\times\mathbb{Z}_2$. One can think of $G$ as the group of symmetries of the icosahedron preserving a pair of opposite points.