Group theory is hard

Let G be a finite group such that G/Z(G)\simeq \Bbb{Z}/p\Bbb{Z}\oplus\Bbb{Z}/p\Bbb{Z}, with p prime. Then p divides the order of Z(G).

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Group theory is hard

One thought on “Group theory is hard

  1. dantegrevino says:

    Let S be a Sylow p-subgroup of G. The order of S is p^r for some number r greater or equal to 2. If S and Z(G) have not trivial intersection, we are done. Otherwise, note that S/Z(G)=G/Z(G) (and r=2). Then SZ(G)=G. But G/Z(G) is abelian, thus for every a,b \in S, the commutator [a,b]=aba^{-1}b^{-1} \in S\bigcap Z(G) = \{e\}. Thus S is abelian and then G is abelian. A contradiction.

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