As written in Abstract Algebra by T. W. Judson:

Lemma 13.4 : Let $G$ be a finite abelian $p-$group and suppose that $g ∈ G$ has maximal order. Then $G$ is isomorphic to $g × H$ for some subgroup $H$ of $G$.

The proof supposes that the reader already knows what maximal order means but I don't know its meaning. I searched internet and I found it either difficult/advanced to understand (e.g.) or irrelevant to specifically its meaning on a group (e.g.).

I am very new to Group Theory. Any clear simple explanation of meaning of maximal order in a group G, would be much appreciated.

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    $\begingroup$ It means it has the largest order of any element of $G$. The notion of a maximal order in ring theory is unrelated. $\endgroup$ – Qiaochu Yuan Oct 16 '15 at 2:08
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    $\begingroup$ @QiaochuYuan - means that if $g_i$ has maximal order in $G$ then order of $g_i$ is greater than (or equal to?) to order of any other elements of $G$, yes? Thank you $\endgroup$ – user231343 Oct 16 '15 at 2:17
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    $\begingroup$ Yes, that's correct. $\endgroup$ – Qiaochu Yuan Oct 16 '15 at 2:22

In a finite group $G$, the order of an element $g\in G$ is the least positive integer $n$ such that $g^n = e$ where $e$ is the 1-element of $G$. You should prove that for each $g$ such exponents $n$ exist (this uses the assumption that $G$ is finite!), so it is possible to pick the least one, so each $g\in G$ has an order.

Now consider all the element orders. Since there are finitely many elements, we just consider a finite set of natural numbers. A finite set of numbers has a maximum. So there exists one or more group elements $g$ whose order is this maximum.

So now you know what is meant by $g$ has maximal order.

(In infinite groups, some elements may have order "infinity" (meaning no positive power of the element is $e$), but if it happens that all elements of an infinite group have a finite order, there may not be a maximum among these element orders.)


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