# $\langle x \rangle$ is a direct summand of a finite abelian group where $x$ is maximal order [duplicate]

This question already has an answer here:

Let $x$ be an element of a finite abelian group $G$ where $x$ has maximal order. Then I want to show that $\langle x\rangle$ is a direct summand of $G$. Note that I do not want to use finite abelian group classification theorem, because I want to use this fact to prove it.

My strategy is to find a subgroup $H\le G$ where $\langle x\rangle\cap H = \{e\}$ and $\langle x,H\rangle = G$. But I'm unsuccessful in finding such $H$.

-

## marked as duplicate by mez, Sami Ben Romdhane, mau, Claude Leibovici, amWhyMay 1 '14 at 11:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I haven't checked the details so I'll leave this as a comment and not an answer. But I think you should use Zorn's lemma. Choose $H$ to maximal among subgroups with the property that $H \cap \langle x\rangle = \{e\}$. Then you'll have to show that this maximality property implies $\langle x, H\rangle = G$, maybe by using $y \in G \setminus \langle x, H\rangle$ to construct a larger $H$. – Jim Apr 30 '14 at 19:33
@Jim Since we are in finite abelian group, we can skip zorn's lemma and just say take a maximal $H$. But first we need to discuss the possibility that $H\cap \langle x\rangle \ne \{e\}$ for all $H$. – mez Apr 30 '14 at 19:38
Ha, silly me. I'm not used to finite things. Anyway, in that case your maximal $H$ would be $H = \{e\}$, so if you can show that being maximal implies $\langle x, H\rangle = G$ then you don't need to worry about this case. – Jim Apr 30 '14 at 19:40

## 2 Answers

Assume $G$ is $p$-primary. Suppose $x$ is of maximal order, a power of $p$. We may assume $G$ is not cyclic. Thus, there exist two proper subgroups of the same order $p$, $H,K$. Let $H$ be the subgroup that is not contained in $\langle x\rangle$ ($\langle x\rangle$ has a unique subgroup of order $p$). It follows $H\cap \langle x\rangle$ is a proper subgroup of $H$, so it is trivial. Hence

$$\overline X=\frac{\langle x\rangle+H}H\simeq{\langle x\rangle}$$

Thus $\overline X$ is a cyclic subgroup of $G/H$, and of maximal order (the same order as $x$). Now choose a direct summand by induction on $|G|$ and lift it to a direct summand in $G$, and conclude using that $G=\bigoplus G(p_i)$ for some primes $p_1,\ldots,p_n$, by taking $x=x_1+\cdots+x_n$ where the $x_i$ are the components of $x$ in the sum.

Claim An abelian $p$-group $G$ is cyclic if it has a unique subgroup of order $p$.

Proof Let $a\in G$ be of largest order, say $|a|=\exp G=p^k$. Let $H$ be the unique subgroup of order $p$. Then $H\leqslant \langle a \rangle$. Suppose there is $x\in G\smallsetminus \langle a\rangle$. We may assume that $x^p\in\langle a\rangle$, since iterating the situation $x^p\notin\langle a\rangle$, we may look at $x^{p^2}$ and so on. In some moment we will reach $x^{p^k}=1\in \langle a \rangle$. Then $x^p=a^m$. If $k=1$ then $x^p=1$ so $x\in H$ which cannot be. Suppose $k>1$. Then $1=x^{p^k}=a^{mp^{k-1}}$, so $p\mid m$. But then $x^p=a^{pl}$, so $x^{-1}a^l$ has order $p$, and is in $H$; which gives $x^{-1}$ is in $\langle a\rangle$, which cannot be. $\blacktriangleleft$

-

We may of course assume that $\langle x \rangle$ is a proper subgroup of $G$.

Suppose there is an element of prime order $y \notin \langle x \rangle$. Then consider $G / \langle y \rangle$ and proceed by induction.

If there is no such $y$, take any $z \in G \setminus \langle x \rangle$. We may assume that $z$ has order a power of a prime $p$, and that $z^{p} \in \langle x \rangle$. If $z^{p} = u^{p}$ for some $u \in \langle x \rangle$, then $z u^{-1}$ is an element of order $p$ not in $\langle x \rangle$, a contradiction.

Thus the $p$-part of the order of $x$ equals $\lvert z^{p} \rvert$, so that $x = a b$, with $a$ of order prime to $p$, and $\lvert b \rvert = \lvert z^{p} \rvert$. One sees now that $a z$ has order $\lvert x \rvert \cdot p$, a contradiction.

-
@PedroTamaroff, it is indeed a misprint, which I have just fixed. Thanks. – Andreas Caranti Apr 30 '14 at 20:04
No problem. Did you see my answer? I am trying to see if there is some similarity between it and this. – Pedro Tamaroff Apr 30 '14 at 20:08
@PedroTamaroff, you go through the details of my second line, which is probably the right thing to do. With another argument, though, it seems I am expanding on yours. Let me explain, how do you show that if $G$ is not cyclic, then there is a prime $p$ such that $G$ has two distinct subgroups of order $p$? I might be just plainly wrong, but there might be a vicious circle of sorts here. – Andreas Caranti May 1 '14 at 12:03
My proof that an abelian $p$-group is cyclic iff it has a unique subgroup of order $p$ follows your idea. – Pedro Tamaroff May 1 '14 at 17:36
@PedroTamaroff, ok! – Andreas Caranti May 1 '14 at 17:37