Groups of Prime Power Order

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If $p$ is a prime, and $G$ is a finite non-cyclic $p$-group, show that there is a normal subgroup $N$ of $G$ such that $G/N$ is isomorphic to $\mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}$.

 

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Spoiler

Proof by induction on $k$, where $|G| = p^k$. The assertion is vacuous when $k = 1$ since every group of prime order is cyclic. Now assume $k > 1$ and the assertion holds for all $p$-groups of order smaller than $|G|$.

If $G$ is abelian, then by the fundamental theorem of finitely generated abelian groups, there is an isomorphism $h$ from $G$ onto a direct sum of cyclic $p$-groups $\mathbb{Z}/p^{a_1}\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/p^{a_m}\mathbb{Z}$. Note $m > 1$ since $G$ is non-cyclic. If $N$ is the preimage of $$p\mathbb{Z}/p^{a_1}\mathbb{Z} \oplus p\mathbb{Z}/p^{a_2}\mathbb{Z} \oplus \bigoplus_{2 < j \le m} \mathbb{Z}/p^{a_j} \mathbb{Z}$$ under $h$, then $N$ is a normal subgroup of $G$ such that $G/N \simeq \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}$.

In case $G$ is nonabelian, $G/Z(G)$ is noncylic. Since the center of a $p$-group is nontrivial, $G/Z(G)$ has order smaller than $|G|$. By the induction hypothesis, there is a normal subgroup $\overline{N}$ of $G/Z(G)$ such that $(G/Z(G))/N \simeq \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}$. By the lattice isomorphism theorem, $\overline{N} = N/Z(G)$ for some normal subgroup $N$ of $G$ containing $Z(G)$. Finally, the third isomorphism theorem yields $G/N \simeq (G/Z(G))/\overline{N} \simeq \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}$.