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Does every finite group $G$ have a set of generators such that the sum of the orders of the generators is less than or equal to $|G|$?

This is surely true but I am failing to see why.

This is easily seen to be true for the symmetric group, abelian groups, the quaternions, and it follows from their classification that simple groups have this property too.

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    $\begingroup$ @Shaun a rare good edit thank you. $\endgroup$ Jan 31, 2021 at 16:46
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    $\begingroup$ You're welcome, @JPMcCarthy :) $\endgroup$
    – Shaun
    Jan 31, 2021 at 16:47

1 Answer 1

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We prove this by induction on $|G|$, the base case being the trivial group. Now suppose $G$ is nontrivial and let $H\subset G$ be a maximal proper subgroup. By the induction hypothesis, there is a set $S$ of generators for $H$ such that the sum of the orders of elements of $S$ is at most $|H|$. By maximality of $H$, $G$ is generated by $H$ together with one more element $x\in G$. If $x$ has order $|G|$, then it generates $G$ and we can take $\{x\}$ as our set of generators for $G$. Otherwise, the order of $x$ and $|H|$ are both proper divisors of $|G|$, and in particular they are both at most $|G|/2$. So their sum is at most $|G|$, and so the sum of the orders of elements of $S\cup\{x\}$ is at most $|G|$.

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    $\begingroup$ That's a lovely proof thank you. $\endgroup$ Jan 31, 2021 at 16:43

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