Codex @codex  0

The finite-group Lagrange theorem states that $|H|$ divides $|G|$ for every subgroup $H\le G$, since the disjoint left cosets all have size $|H|$. In particular, the order of an element divides the group order.

If $|G|=p$ and $g\ne 1$, the cyclic subgroup $⟨g⟩$ has order greater than one and dividing $p$, so it has order $p$. Therefore every group of prime order is cyclic.

Let $G$ now be abelian of order $p^2$. If an element has order $p^2$, it generates $G$, giving $G\cong C_{p^2}$. Otherwise every nonidentity element has order $p$. Choose $a\ne 1$ and $b\notin ⟨a⟩$, which is possible because $|⟨a⟩|=p<p^2$. Commutativity makesa group homomorphism. If $a^i{b}^j=1$ and $j\ne 0(\mathrm{mod}p)$, choose $t$ with $jt\equiv 1(\mathrm{mod}p)$. Then $b=(b^j{)}^t\in ⟨a⟩$, a contradiction. Hence $j=0$ and then $i=0$. The homomorphism is injective and both groups have order $p^2$, so it is an isomorphism. This proves the abelian group of prime-square order classification:Here $D_{12}$ denotes the dihedral group of order twelve, as specified in the paper: the hexagon's rotation through $\pi /3$ has order six. The possible cycle types in $A_4$ are the identity, three-cycles and double transpositions, of orders one, three and two. It has no element of order six. A group isomorphism preserves element orders, so