# What is the largest order of element in an abelian group of order $5!$?

What is the largest order of element in an abelian group of order $5!$ ?

I don't know how to deal with problem. Can anyone give a solution or useful resources?

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What is an "associate group 5!"? Or is the exclamation mark part of the sentence? – Phira Nov 3 '12 at 15:34
It's not exclamation. I have modified the question. – John Hass Nov 3 '12 at 15:38
E.g. In the cyclic (thus abelian) group $\mathbb{Z}_{5!}$, any prime number $p \text{ such that}\; p \in \mathbb{Z}_{5!}, p \neq 2, p<5!,$ has order 5! – amWhy Nov 3 '12 at 15:54

Well, $5!=2^3\cdot3\cdot5$, so by the fundamental theorem of finitely generated abelian groups, there are $3$ abelian groups of order $5$:

$$\mathbb{Z}_{2^3}\times \mathbb{Z}_{3} \times \mathbb{Z}_5$$ $$\mathbb{Z}_{2^2}\times \mathbb{Z}_2 \times \mathbb{Z}_{3} \times \mathbb{Z}_5$$ $$\mathbb{Z}_{2}\times \mathbb{Z}_2\times \mathbb{Z}_2 \times \mathbb{Z}_{3} \times \mathbb{Z}_5$$

The first one is isomorphic to $\mathbb{Z}_{5!}$ so of course the maximum order of an element is $5!=120$.

The second one is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_{3\cdot 4 \cdot 5}$ so the maximum order of an element is $3\cdot 4 \cdot 5=60$.

And the third one is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_{3\cdot 2 \cdot 5}$ so the maximum order of an element is $3\cdot 2 \cdot 5=30$.

So if your question is, "What is the largest order of an element in a given abelian group of order $5!$?" then the answer is one of the three above, depending on which group you were given. On the other hand, so if your question is, "What is the largest order of an element in any abelian group of order $5!$?" then the answer is clearly $120$, and furthermore, in general, for any $n\in \mathbb{N}$ the maximum order of an element in any group of order $n$ is $n$.

If you don't know the fundamental theorem of finitely generated abelian groups, you should look that up - it's very useful. Another thing you may want to do is look at this post about using the least common multiple to find element orders of abelian groups, which would provide an alternative path to this solution.

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A very late thanks! – John Hass Nov 8 '12 at 17:18

Assuming that you are asking "what is the largest order for an element in a group of order 5! ?", the answer is 5!. Just take a cyclic group of order 5!.

If that is not what you meant, please explain a bit further.

If, on the other hand, you want the largest order for an element in a non-abelian group of order 5!, then that would be $5!/2$, by considering a dihedral group of order $5!/2$, which has a cyclic subgroup of order $5!/2$, and hence an element of that order.

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updated. It's Abelian group. – John Hass Nov 3 '12 at 15:43
Yes, a cyclic group IS abelian. – Old John Nov 3 '12 at 15:44
@OldJohn : ) ${}{}$ – Rudy the Reindeer Nov 3 '12 at 16:15