In Serge Lang’s Algebra he leaves a direct proof of the following statement to the reader:

Let $G$ be a finite abelian group. If $G$ is not cyclic, then there exists a prime $p$ and a subgroup of $G$ isomorphic to $C$$\times$$C$, where $C$ is cyclic of order $p$.

I have come up with this so far:

Consider such a group $G$ and let $r$,$s$ be non-identity elements such that neither can be expressed as a power of the other (such elements exist because $G$ is not cyclic.) We then have two cases: either the periods of $r$ and $s$ are coprime or they aren't (and thus both are divisible by some prime $p$.) Suppose the periods are $m$ and $k$ respectively.

  1. If their periods are coprime, then let $p$ divide $mk$. The group $\langle r\rangle\times\langle s\rangle$ is then cyclic(since $m$ and $k$ are coprime) of order $mk$. Given a generator $x$, the cyclic group $\langle x^{mk/p}\rangle$ is of order $p$ and we have our group $C$. Now we must prove that $s^i$ does not equal $r^k$ for any nontrivial $i,k$. If this were true, then $r^k$ would generate a subgroup of $\langle s\rangle$. The order of this subgroup then divides both $\langle s\rangle$ and $\langle r\rangle$ contradicting $m$ and $k$ being coprime. Hence we have an isomorphism between the subgroup $\langle s,r\rangle$ and $\langle x^{mk/p}\rangle$.

  2. I can not seem to find a similar result for the case where $m$ and $k$ are not coprime (that is, I cannot figure out why $(r^{m/p})^i=(s^{k/p})^j$ cannot happen for nontrivial $i,j$.)

So, overall, is what I have in the first case correct and how do I prove the second?

  • 1
    $\begingroup$ Please observe how I have edited your $\rm\LaTeX$ code. There is no need to type $m$$k$ for $mk$; instead, type $mk$. $\endgroup$
    – Shaun
    Commented Jul 11, 2021 at 14:10
  • $\begingroup$ You won't be able to prove the parenthetical comment in case $2$, because that conclusion need not hold. For example, consider the group $C_2\times C_4$ (cyclic order $2$ times cyclic order $4$, generated by $x$ and $y$, respectively) and say $r=(1,y)$, $s=(x,y)$, both of order $4$. Then $r^{4/2} = (1,y^2) = s^{4/2}$. $\endgroup$ Commented Jul 14, 2021 at 3:35
  • $\begingroup$ Suggestion: start with an element of maximal possible order for $r$. Divide your cases into what happens when the two elements generate cyclic subgroups that intersect trivially (proceed as you do in case 1), and when they don't. When they don't, try to tweak one of them so that they now generate subgroups that intersect trivially. $\endgroup$ Commented Jul 14, 2021 at 3:36

2 Answers 2


Here's a proof that does not use the structure theorem for finite(ly generated) abelian groups. I will use multiplicative notation throughout

Let $h\in G$ be of maximal order, and let $H=\langle h\rangle$; let the order of $h$ be $n$. We know that $H\neq G$, since $G$ is not cyclic. Let $x\in G$ be such that $x\notin H$. Let the order of $x$ be $m$.

I claim that $m$ must divide $n$. Indeed, in an abelian group, the orders of elements are lcm-closed; that is, if you have an element of order $m$ and an element of order $n$, then there must exist an element of order $\mathrm{lcm}(m,n)$. Since $n$ is the largest possible order, we must have $\mathrm{lcm}(m,n)\leq n$, and therefore $\mathrm{lcm}(m,n)=n$, proving that $m$ divides $n$, as claimed.

Case 1. $\langle x\rangle \cap H=\{e\}$.

Let $p$ be a prime that divides $m$. Then $x^{m/p}$ has order $p$, $h^{n/p}$ has order $p$, and $\langle x^{m/p}\rangle\cap\langle h^{n/p}\rangle=\{e\}$, so $\langle x^{m/p},h^{m/p}\rangle \cong C_p\times C_p$, as desired.

Case 2. $\langle x\rangle \cap H\neq \{e\}$.

Say $r\gt 0$ is the smallest integer such that $x^r\in H$. Because $x\notin H$, we know that $r\gt 1$. Since $\langle x^r\rangle$ is a proper subgroup of $\langle x\rangle$, we know that $r|n$. Let $s$ be the smallest positive integer such that $h^s\in\langle x\rangle\cap H$. Then we also know that $s|n$, and $s\gt 1$. Moreover, $\langle x^r\rangle = \langle h^s\rangle$, so the order of $h^s$ is the same as the order of $x^r$. That is, $\frac{n}{s}=\frac{m}{r}$. Therefore, $s = \left(\frac{n}{m}\right)r$, and since $m$ divides $n$, it follows that $r$ divides $s$. Write $s=rk$.

Now, $x^r\in \langle h^s\rangle$, so there exists $t$ such that $x^r=(h^s)^t = h^{st} = (h^{kt})^r$. Consider $y=xh^{-kt}$; note that $y\notin\langle h\rangle$, since $x\notin\langle h\rangle$. Also, the smallest positive integer $c$ such that $y^c\in\langle h\rangle$ is $c=r$, because if $y^c\in\langle h\rangle$, then $x^ch^{-ktc}\in \langle h\rangle$, hence $x^c\in\langle h\rangle$. But $y^r = x^rh^{-ktr} = x^r h^{-st} = e$. Thus, $\langle y\rangle \cap \langle h\rangle = \{e\}$, and we are reduced to Case 1, proving that $\langle y,h\rangle$ contains a subgroup of the form $C_p\times C_p$, as desired.

  • $\begingroup$ Does it follow as a corollary that, if $n$ is even, then $p=2$ fulfils the claim? Equivalently, is there always, for even $n$, an $x\in G\setminus H$ of even order? $\endgroup$
    – Kan't
    Commented Nov 16, 2022 at 12:50
  • 1
    $\begingroup$ @Devo: It's not a corollary, since I do not control $m$, and I could pick an $m$ of odd order (e.g., if the group is $\mathbb{Z}_3\times\mathbb{Z}_6$, I take $(0,1)$ as $h$, and $(1,0)$ as $h$). But if the largest order is even, then there is always an element of even order: if $x$ is of odd order, then $xh$ will necessarily have even order. $\endgroup$ Commented Nov 16, 2022 at 15:26
  • $\begingroup$ What is $g$ (lowercase) in the second paragraph? $\endgroup$
    – Kan't
    Commented Sep 8, 2023 at 21:45
  • $\begingroup$ @citadel a typo for $h$. $\endgroup$ Commented Sep 8, 2023 at 21:51

Sorry I don’t have enough reputation to comment. But I think here it is good to use the fundamental theorem of finitely generated abelian groups. You know $G$ is a product of cyclic groups. $G \cong \mathbb{Z}_{n_1} \times \dots \times \mathbb{Z}_{n_k}$ Also since $G$ is not cyclic two of the cyclic groups in the product have orders which are not coprime. For simplicity say we have a prime $p$ so that $p$ divides both $n_1$ and $n_2$. Can you use this to find a subgroup of G isomorphic to $\mathbb{Z}_{p} \times \mathbb{Z}_{p}$ ?

I think I came up with something else but it uses Sylow’s theorem. Suppose that $G$ is finite and abelian with order $n=p_1^{r_1} \dots p_k^{r_k}$ where the $p_i$ are distinct primes. For each $i$ let $H_i$ be a Sylow $p_i$-subgroup of $G$ which exists by Sylow’s theorem. Note that not all of these subgroups can be cyclic. Indeed if this were the case there would be elements $x_i \in H_i$ of order $p_i^{r_i}$ but then one can check that there product $$x= x_1 x_2 \dots x_k$$ would have order $n$ contradicting that $G$ is not cyclic. Now we can denote by $H$ a noncyclic Sylow $p$-subgroup of order $p^r$. As you noted in your first post this means there are elements $g$ and $h$ such that neither is a power of the other. If $g$ and $h$ are of orders $p^{r_1}$ and $p^{r_2}$ respectively then $g^{p^{r_1-1}}$ and $h^{p^{r_2-1}}$ both have order $p$ and generate a subgroup of $G$ isomorphic to $\mathbb{Z}_p \times \mathbb{Z}_p$

  • $\begingroup$ That makes sense, however the textbook hasn’t covered the fundamental theorem of finitely generated abelian groups. Thank you regardless! $\endgroup$ Commented Jul 11, 2021 at 14:57
  • $\begingroup$ Ah I see! That is a little strange, what page of Lang does he mention this exercise? $\endgroup$
    – Hankry
    Commented Jul 11, 2021 at 18:01
  • $\begingroup$ Page 25. It’s part 6 of proposition 4.3. $\endgroup$ Commented Jul 11, 2021 at 19:43
  • $\begingroup$ Oh! I see what you mean, he does say in the text that this result should be easier to prove than the structure theorem, sorry I misunderstood, I’ll try to think of a different proof $\endgroup$
    – Hankry
    Commented Jul 11, 2021 at 19:55

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