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For weeks I tried to solve the following question on Brilliant: Fill in the blank: "Every group of order ___ is abelian." And these are the possible answers I get: 15, 16, 20, 21, 27. Using Sylow's theorems I tried my best to get to a conclusive answer, but unfortunately all I could do is guess that it is probably 15, because the others seem unlikely. So I clicked on 15 and submitted my answer and it turns out that my gut feeling was right. Hooray! But what was the explanation?

Obviously a group G of order 15 has Sylow subgroups of orders 3 and 5. It's also quite obvious they have to be normal in G, because of Sylow's third theorem. This means that their product is isomorphic to G. And since they are both abelian, so is G.

Hang on?! Any group of order 3 obviously must be abelian, but how was I supposed to know that a Sylow 5-subgroup is abelian as well? It is possible for a group of order 5 to be non-abelian, so what piece of information am I missing here?

Are Sylow 5-subgroups always abelian and if so, how was I supposed to know that?

Shaun
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user3635700
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    Any group of prime order is cyclic and thus Abelian – J. W. Tanner Sep 26 '23 at 20:07
  • Incidentally, for an elimination approach: I can think of $He_2 \times C_2$ of order 16; a semidirect product induced by $C_4 \simeq \operatorname{Aut}(C_5)$ of order 20; a semidirect product induced by $2 \cdot : C_3 \to \operatorname{Aut}(C_7) \simeq C_6$ of order 21; and $He_3$ of order 27. – Daniel Schepler Sep 26 '23 at 21:39
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    Wondering how you got to Sylow theorems without covering Lagrange's theorem... – Dustan Levenstein Sep 26 '23 at 21:45
  • A remark to your task :"Every group of order ___ is abelian." The blank cannot be filled with an even number bigger than $4$, because we always have a non-abelian group $D_n$ of order $2n$ for $n\ge 3$. So $16$ and $20$ are impossible. A similar argument works for $n=21$ and $27$. – Dietrich Burde Sep 27 '23 at 15:02

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It is possible for a group of order 5 to be non-abelian, so what piece of information am I missing here?

No, it is not possible.

Any group of prime order is cyclic (as shown here) and thus (as shown here) Abelian.

J. W. Tanner
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Are Sylow 5-subgroups always abelian and if so, how was I supposed to know that?

No.

Consider a nonabelian group $H$ of order $5^r$, some $r$. Then $H$ is a (really, the) Sylow $5$-subgroup of

$$G=H\times\Bbb Z_2.$$

Let $H={\rm He}_5$ for example.

Shaun
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