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I had a test in representation theory today and one of the questions regarded a non-abelian group of order 725. However, I think I managed to prove that every group of order 725 is abelian. If I have a mistake in the proof, I will be happy if you answer about it.

Suppose $G$ is a group of order 725. We have $725=5^2\cdot29$ so the dividers of 725 are $1,5,25,29,145,725$. Let $s$ be the number of sylow 5-subgroups of $G$. According to sylow theorems, $s\equiv1\mod{5}$ and $s\mid725$ so the only option is $s=1$ and $G$ has one normal sylow 5-subgroup, we will denote it by $P$.

The group $G/P$ has 29 elements so it is cyclic, so $G'\subseteq P$, so $\vert G'\vert=25$ or $\vert G'\vert=5$ or $\vert G'\vert=1$.

For every $d\in\mathbb{N}$, let $n_d$ be the number of represtations of degree $d$ of $G$ up to equivalence. Then $n_d=0$ for every $d$ such that $d\nmid725$. Also, we have $d^2n_d\leq725$ for all $d$, and $29^2>725$ so $n_{29}=n_{145}=n_{725}=0$, so

$n_1+25n_5+625n_{25}=725$

So $25\mid n_1$. But $n_1$ is equal to the number of one-dimensional representations of $G$, which is the number of one-dimensional representations of $G/G'$, which is $\vert G/G'\vert$. So 25 divides $\vert G/G'\vert$, and it is only possible if $\vert G'\vert=1$, so $G$ is abelian.

Shaun
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uri gluck
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  • Are you trying to prove that any group (regardless of the operation) of order 725 is abelian? – MathGeek Oct 04 '21 at 17:42
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    Yes, and it is true for some numbers that every group of that order is abelian (for instance every group of a prime order is abelian) – uri gluck Oct 04 '21 at 17:45
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    See https://oeis.org/A051532 and https://math.stackexchange.com/questions/1556811/does-there-exist-an-n-such-that-all-groups-of-order-n-are-abelian – lhf Oct 04 '21 at 17:52
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    I think your proof is OK, but you don't really need any representation theory for this question. You proved that there is a unique and therefore normal Sylow $5$-subgroup $P$, and similarly you can prove that there is a unique Sylow $29$-subgroup $Q$. Then $P$ and $Q$ are both abelian, and $G = P \times Q$, so $G$ is abelian. – Derek Holt Oct 04 '21 at 18:44
  • Your write-up introduces a subgroup $G'$ without explicit definition. Defining it at that point, perhaps with the stated goal of showing it to be the trivial subgroup as you subsequently did, would improve clarity. – hardmath Oct 04 '21 at 22:10

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