Let $G$ be a nonabelian finite 2-generated (ie, can be generated by 2 elements) $p$-group.
Must its Frattini quotient be $C_p\times C_p$, where $C_p$ is the cyclic group of order $p$? (Equivalently, its abelianization is not cyclic?)
abelianization of 2-generated finite $p$-groups
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abstract-algebra
group-theory
finite-groups
1 Answers
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Suppose $G$ is a $p$-group generated by $x$ and $y$, let $f:G\to H$ be the Frattini quotient, and suppose $H$ is not isomorphic to $C_p\times C_p$. Since $H$ is generated by $f(x)$ and $f(y)$, it must then be cyclic, and so it is actually generated by just one of $f(x)$ and $f(y)$; say $H$ is generated by $f(x)$. Then we have $f(y)=f(x)^n$ for some $n\in\mathbb{Z}$. Defining $y'=yx^{-n}$, $G$ is generated by $x$ and $y'$ and now $f(y')=1$. But this means $y'$ is a non-generator of $G$, so $G$ is actually generated by just $x$. So $G$ is cyclic and hence abelian.
More generally, a similar argument shows that if $G$ is a $p$-group whose Frattini quotient has dimension $d$, then $G$ can be generated by $d$ elements.