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If $G$ is an abelian group of order $p^n$, $p$ a prime and $n_1\geq n_2\geq \cdots \geq n_k > 0$, are the invariants of $G$, show that the maximal order of any element in $G$ is $p^{n_1}$.

In my self study class my TA told me not to worry about this problem and do not look at it because we are not going to learn it. I was still wondering how this proof would look like because it seems like a good proof to learn for other problems that I might solve later on.

I understand the definitions by some of the things I read and what some other people told me. Is there a way you can prove this though? I would like to see it.

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By the definition of invariants of $G$, it can be decomposed as a product of cyclic groups of orders $p^{n_1},p^{n_2},\ldots,p^{n_k}$. The first factor already contains an element of order $p^{n_1}$, so you are sure this order occurs in $G$. Now directly form the definition, the order of an element in a product of groups$~G_i$ is the least common multiple of the orders of its components in the factors$~G_i$. Here you can easily see that the orders of those components always divide $p^{n_1}$, so the maximal value for that least common multiple is $p^{n_1}$ (and all other orders that occur divide this number).

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I assume your class covered the classification of finitely generated abelian groups. In your case it implies $$G=\oplus \mathbb{Z}/p^{n^{i}}\mathbb{Z}$$here $n^{i}\ge n^{j}$ for $i\ge j$. Then any element can be written in the form $$(..a_{i},...a_{j}...),a_{i}\in \mathbb{Z}/p^{n^{i}}\mathbb{Z}$$

It is clear now that its order is limited by the highest ranking coordinate. In our case every element has order at most $p^{n^{k}}$, with $k$ being the highest indexed coordinate.

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  • $\begingroup$ Yes. That is what I was getting at. That is why I put that G was an abelian group. I just needed to put this together. $\endgroup$ – 9959 Apr 24 '13 at 6:48

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