Classification of Groups of Order p^2

finite-groupsgroup-theoryproof-verification

The book I am reading contains contains the following claim:

Every group of order $p^2$, where $p$ is a prime, is isomorphic to $\mathbf{Z}_{p^2}$ or $\mathbf{Z}_p \oplus \mathbf{Z}_p$.

I tried to do the proof myself and ended up with the following. Is the following sketch correct? —

  1. We can show all groups of order $p^2$ are abelian by examining the center of the group. https://proofwiki.org/wiki/Group_of_Order_Prime_Squared_is_Abelian

  2. We are left to show, that such an abelian group of order $p^2$ is of the two forms given above. Examine arbitrary subgroup $\langle a \rangle \subset G$. By lagrange's theorem it's order can be $p^2$, $p$, or $1$.

    1. If $|\langle a \rangle|=p^2$ we obtain an abelian subgroup order $p^2$ — i.e. $\langle a \rangle$ generates G. So $\mathbf{Z}_{p^2} = \langle a \rangle = G$ (the first isomorphism established by sending $k \mapsto a^k$).

    2. If $|\langle a \rangle|=p$ we have an abelian subgroup of $G$ order of $p$. We can see $\mathbf{Z}_{p} = \langle a \rangle$ by again sending $k \mapsto a^k$. Construct $G/\langle a \rangle$ (Can check $\langle a \rangle$ is normal in $G$). $|G/\langle a \rangle| = p$ by using coset definition and counting. By using Lagrange's theorem, one can show any group of order $p$ is cyclic and abelian. Thus we can construct the isomorphism: $\mathbf{Z}_{p} = G/\langle a \rangle$. We note: $\langle a \rangle = \mathbf{Z}_{p} = G/\langle a \rangle$. And so: $G/\langle a \rangle \oplus \langle a \rangle=\mathbf{Z}_p \oplus \mathbf{Z}_p$. (How do I show $G/\langle a \rangle \oplus \langle a \rangle$ = $G$? I think this is still missing)

    3. If $|\langle a \rangle|=1$ then $G/\langle a \rangle =G$. So we can refer to 1 and 2.

Best Answer

Hint: By Lagrange's every element in $G $ will have order either $1,p$ or $p^2$.

Case-1: If some $g\in G$ has order $p^2$ then $G\cong \Bbb Z_{p^{2}}$.

Case-2: Suppose each non-identity element has order $p$. Let $e\neq a\in G$. Then $\langle a\rangle<G$ of order $p$. Also $\exists$ (why) $b\in G$ such that $b\notin \langle a\rangle$. Next show that G=$\langle a\rangle \oplus\langle b\rangle$ and finally give an isomorphism between G and $\Bbb Z_p \oplus\Bbb Z_p$.

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