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I found this exercise

Prove that any Abelian group of order $45$ has an element of order $15$. Does every Abelian group of order $45$ have an element of order $9$?

In the answer to this last question I am told no. Since $\mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_5$, does not have an element of order $9$.

This is a bit confusing to me, since if $m$ divide order of a finite Abelian group $G$, then $G$ has a subgroup of order $m$ and in this case $9\mid 45$. What detail am I missing?

Hopmaths
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  • https://math.stackexchange.com/questions/108405/explicit-descriptions-of-groups-of-order-45 – stoic-santiago Mar 22 '21 at 16:48
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    'subgroup of order $m$' here means that the size of the subgroup is $m$. You're confusing the order of the group with the order of an element (which, to be fair, is a completely understandable confusion and a painful overlap of notation.) A group of order $n$ doesn't necessarily have an element of order $n$; here the $C_3\times C_3$ factor ($C_n$ being the cyclic group of order $n$; $\mathbb{Z}_n$ is somewhat overloaded notation too) has order $9$, but no element of $C_3\times C_3$ has order $9$. – Steven Stadnicki Mar 22 '21 at 16:52

1 Answers1

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There's a difference between having a subgroup of order $9$ and an element of order $9$. While it's true that every abelian group of order $n$ has a subgroup of order $m$ for any $m|n$, this doesn't necessarily guarantee the existence of an element $x$ of order $m$; it's possible that (as in your case) the subgroup of order $m$ isn't cyclic, and so has no element with the same order as the order of the subgroup.

For concreteness, in your example of $\mathbb Z_3\times\mathbb Z_3\times \mathbb Z_5$, the subgroup $\mathbb Z_3\times\mathbb Z_3$ consisting of the first two factors is of order $9$ (it has $9$ elements), but all of its non-identity elements have order $3$ (have $x^3$ equal to the identity).

Carl Schildkraut
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  • Oh I understand Mr. Carl, thank you. I thought about this but didn't know how to answer it. His explanation and Randall's example have made everything clear to me. If I may abuse a little, could you tell me in this case what a subgroup of order 9 would be? – Hopmaths Mar 22 '21 at 17:00
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    @Hopmaths I've updated my answer. – Carl Schildkraut Mar 22 '21 at 17:02