Vtyjkyuk

We know that $Z(G)<G,;$ then $O(Z(G)) mid O(G). $

If $;O(Z(G))= p^2, $ then $;Z(G)=G$ and we are done.

Now, if $O(Z(G))= p,,$ how can I prove that $G$ is abelian ?

Is it by proving that $G/Z(G)$ is cyclic? And if so, then $G$ is abelian.

If yes how to prove that that $G/Z(G)$ is cyclic?

Prove the following (easy):

Lemma: For any group $;G;$, the quotient group $;G/Z(G);$ cannot be cyclic non-trivial.

From the above it follows that $;G/Z(G);$ cyclic $;implies G;$ is abelian.

$o(Z(G))=p$ then $o(G/Z(G))=p$ and every group of prime order is cyclic

First note that $Z(G)$ is non-trivial, by the class equation. Hence $Z(G)$ has order $p$ or $p^2$. If $O(Z(G))=p^2$, $Z(G)=G$ and we're done.

If $O(Z(G))=p$, then $Z(G)$ is cyclic, generated by, say, $z$. G/$Z(G)$ also has order p, and hence is cyclic, generated by, say $a$. Now let $g,g'$ two elements of $G$. These can be written as $g=a^n z$, $g'=a^n'z',enspace z,z'in Z(G)$. Then
beginalign*gg'&=(a^nz)(a^n'z')=a^na^n'z'z&&textsince $zin Z(G)$\
&=a^n'a^nz'z=a^n'z'a^nz=g'g.&&textsince $z'in Z(G)$endalign*
Thus $G$ is abelian in that case too.

Z(G) is always subgroup of group G therefore by Sylow's first theorem the possibility for

$o(Z(G))=1 or p or p^2$ but p groups have nontrivial center thus,$o(Z(G))= p or
p^2$ if $o(Z(G))=o(G)$ then G is Abelian(can you prove this?).

Now, Using this lemma if $o(Z(G))=p$

$;G/Z(G)=;$ cyclic $;implies G;$ is Abelian.

$;o(G/Z(G))=p;$ hence it is cyclic implies that G is Abelian.

: Let P be a group of order $p^2$
. We know that every p−group has
non-trivial, so the center $Z(P)$ of $P$ is not trivial.Thus $|Z(P)|$ is either $p$ or $p^2$.In
the latter case, we have $P = Z(P)$, so P is abelian. In the former case, $P/Z(P) $is
a group of order p. We know that groups of prime order p are cyclic, so $P/Z(P)$ is
cyclic. P is abelian so $ Z(P) = P,$ a contradiction