Group $G$ is abelian and finite. $\langle g\rangle = G$. $p$ is order of $G$ (and $\langle g\rangle$). $p=mn$, $m > 1$, $n > 1$. Why $\langle g^m\rangle < G$ (not $\langle g^m\rangle \le G$)?
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3You are using $p$ for a composite number! – Angina Seng Dec 25 '19 at 03:28
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$\left<g^m\right>=G$ iff $m=1$. – Angina Seng Dec 25 '19 at 03:29
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$g^p=(g^m)^n=e$, so $\lvert\langle g^m\rangle\rvert\le n<mn=p=\lvert G\rvert $ – J. W. Tanner Dec 25 '19 at 03:32
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1Note that “$\langle g^m\rangle\lt G$” implies $\langle g^m\rangle \leq G$. They are both correct statements. It’s just that the former is stronger than the latter because it says more. – Arturo Magidin Dec 25 '19 at 03:39
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if $G=\langle g\rangle$ then group $G$ is cyclic so certainly abelian – J. W. Tanner Dec 25 '19 at 03:42
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Basic result in theory of cyclic groups: $\langle g\rangle =G\implies |g^m|=\dfrac n{\operatorname {gcd}(n,m)}$, where $n=|G|$.
Angina Seng
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As $|\langle g^m \rangle | = \dfrac {|G|} m$ and $m > 1$, so $|\langle g^m \rangle | < |G|$.
