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I'm trying to prove that a finite abelian group $G$ is generated by elements of maximal order.

I can sort of see why that happens in a vague instinctive kind of way but no real hard logic. So far, I have tried to use this lemma:

Let $G$ be a finite abelian group and a be an element of maximal order in $G$. Then any element $b$ is such that $|b|$ divides $|a|$.

Then each cyclic group generated by the maximal order element contains exactly 1 element of each possible order for an element in the group. I have tried expanding from this idea in a few directions but I don't think it's the correct way to go since it's not going anywhere.

Any help appreciated. Thanks guys.

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  • $\begingroup$ What does it mean for $a$ to have maximal order? If there is no $b$ such that $a$ is in group generated by $b$ and $b$ has order greater than the order of $a$? Then the result is immediate. Or do you mean nothing in the group has order bigger than the order of $a$? $\endgroup$ Jan 5, 2013 at 21:50
  • $\begingroup$ I would imagine $a$ having maximal order to mean $a$ has the largest order possible among all elements of $G$. Would this be the same as your condition? $\endgroup$ Jan 5, 2013 at 21:53
  • $\begingroup$ Not the same as the first. Thanks, now I understand the problem. $\endgroup$ Jan 5, 2013 at 21:55

5 Answers 5

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Let $G$ be a finite abelian group, then $G = \mathbb{Z}_{n_1}\oplus\cdots\oplus \mathbb{Z}_{n_k}$ for some $n_1|\cdots|n_k$. Then clearly these elements $$\{(1,\cdots,1),(0,1,\cdots,1),(0,0,1,\cdots,1),\cdots,(0,\cdots,0,1)\}$$ are of maximal order $n_k$ and they generate $G$.

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The exponent of a finite group $G$ is the smallest positive integer $e$ such that $x^{e} = 1$ for all $x \in G.$ When $G$ is Abelian, this is also the maximal order of an element of $G$, and does occur as the order of an element, as you seem to know, since it's equivalent to the Lemma you mention. Now consider the case that $G$ is Abelian of exponent $e,$ and let $y$ be an element of $G$ of order $e.$ Let $\{x_{1},x_{2},\ldots x_{n} \}$ be a generating set for $G$ consisting of elements of prime power order- such a generating set exists (there may be much smaller generating sets). We have another generating set $\{y,x_{1},\ldots, x_{n} \}.$ If any $x_{i}$ is a power of $y,$ we can remove it from the new generating set, and still be left with a generating set, so we may produce a generating set $\{y,x_{1},\ldots, x_{m} \}$, possibly after relabelling, such that each $x_{i}$ has prime power order, and no $x_{i}$ is a power of $y.$ Let the order of $x_{i}$ be power of a prime $p_{i}$ (the $p_{i}$ need not be distinct).

If $e$, the order of $y$ is a power of a prime $p,$ then each $p_{i} = p.$ In that case, if $x_{i}$ has order less than $e,$ then $x_{i}y$ still has order $e$. So in that case, we obtain a new generating set, consisting all of elements of order $e,$ by incuding $y$, and including each $x_{i}$ of order $e, $ but replacing $x_{i}$ by $x_{i}y$ whenever $x_{i}$ has order less than $e.$ Notice that we can still recover each $x_{i}$ using this generating set.

Suppose then that $e$ is not a prime power. If the order of $x_{i}$ is less than the power of $p_{i}$ dividing $e,$ then we replace $x_{i}$ by $yx_{i},$ which has order $e.$ If the order of $x_{i},$ say $e_{i}$, is equal to the power of $p_{i}$ dividing $e,$ then we replace $x_{i}$ by $x_{i}y^{e_{i}},$ which still has order $e.$ We can still recover each $x_{i}$ from this generating set, so we still have a generating set, but this time, all generators have order $e.$

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    $\begingroup$ It isn't really necessary to treat the case that $y$ has prime power order separately. $\endgroup$ Jan 5, 2013 at 22:44
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By the fundamental theorem of finite abelian groups there exists a decomposition of $G$ as the direct sum $$\bigoplus_{i=1}^{n} \mathbb{Z}/d_i\mathbb{Z}$$ such that $d_1|d_2| ... |d_n.$ Fix such a decomposition and let $H = (\bigoplus_{i=1}^{n-1}\mathbb{Z}/d_i\mathbb{Z})\oplus0.$ The maximal order of an element of $G$ is $d_n.$ If $1$ is a generator of $C := (\bigoplus_{i=1}^{n-1}0)\oplus \mathbb{Z}/d_i\mathbb{Z}$ and $f \in H,$ then $f + 1$ and $1$ both have order $d_n.$ It follows that the subgroup generated by maximal elements contains every element of $H$ and every element of $C$ and thus equals $G.$

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Let $x$ be an element that doesn't have maximal order, and $y$ be any element that has order $n$, which is the maximal order.

If the subgroup $\langle x, y \rangle$ is cyclic group, then its order can't be larger than the order of $y$, and so $x \in \langle y \rangle$.

Otherwise, $\langle x, y \rangle$ is the product of two cyclic groups of order $m,n$. There is an element of order $\operatorname{lcm}(m,n)$, so we have $m | n$. We can pick generators $a,b$ of the two group. Then $(a,b)$ and $(0,b)$ are two elements of maximal order, and $x$ is in the subgroup generated by them.

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let a be an element of a prime order such that is not belong to the subgroup of G generated by all maximal elements then ab is not belong to the set of the generators of maximal elemens for some b of maximal elements.

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    $\begingroup$ Thank you for your answer. Sadly, it is phrased in a way that is incomprehensible to me. Could you please rephrase it and add some more detail? When answering questions this old that already have other answers, anything short of a fully detailed solution is not adding value to the question. $\endgroup$
    – Lord_Farin
    Nov 8, 2013 at 11:48

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