# A finite abelian group that does not contain a subgroup isomorphic to $\mathbb Z_p\oplus\mathbb Z_p$, for any prime $p$, is cyclic.

Suppose that $G$ is a finite abelian group that does not contain a subgroup isomorphic to $\mathbb Z_p\oplus\mathbb Z_p$ for any prime $p$. Prove that $G$ is cyclic.

Attempt: If $G$ is a finite abelian group, then let $H$ be any subgroup of $G$

It's given that $H \not\simeq \mathbb{Z}_p \oplus \mathbb{Z}_p$ which can be due to a variety of reasons like : $|H|$ may not be $p^2$ or $H$ may not contain any element of order $p$ etc

Hence, the process of finding an element $g$ such that $|g| = |G|$ seems difficult, hence, probably the best bet would be to first assume that $G$ is not cyclic.

Hence, $O(g) \neq |G|~ \forall ~ g \in G$ .

Also, $G \not\simeq Z_{|G|}$ since $G$ is not cyclic.

How do i arrive at a contradiction from here that if $G$ is not cyclic, it must contain a subgroup $H \simeq \mathbb Z_p \oplus \mathbb Z_p$.

Please note that this question occurs in Gallian before normal and factor groups are introduced.

Thank you for help.

• $|\mathbb{Z}_p\oplus\mathbb{Z}_p|=p^2,$ not $2p$. – Kevin Carlson Apr 20 '14 at 15:31
• yeah, I corrected it. – MathMan Apr 20 '14 at 15:34
• Let me start with a hint, and let me know if it helps: try induction on the number of terms in the prime factorization of $|G|$. – Kevin Carlson Apr 20 '14 at 15:36
• Then $G$ is "square free": $G \equiv \prod \mathbb Z_{p_i}$ where the $p_i$ are unique primes. By the CRT, $G$ is cyclic and generated by the element $1 \leftrightarrow (1,1,1,\cdots,1)$. – steven gregory May 28 '16 at 14:29

Factor $|G|$ into primes as $\prod_{i=1}^n p_i^{k_i}$ where the $p_i$ are distinct. Proceed by induction on $n$: the base case is done assuming you know prime-order groups are cyclic. Then for the induction step factor out all instances take a generator $g_1$ of a subgroup of order $p_1^{k_1}$ and $g_2$ of a subgroup of order $|G|/p_1^{k_1}$. (By induction such a subgroup is cyclic since if it contained $\mathbb{Z}_p\oplus \mathbb{Z}_p$ so would $G$.)

I claim $g_1+g_2$ is a generator of $G$. For suppose $g_1+g_2$ has order $m$. Then $mg_1=-mg_2$. Now $mg_1$ is of order $p_1^n$ for some $1\leq n\leq k_1$, so $p_1^n mg_2=0$ and $|G|/p_1^{k_1}$ divides $p_1^nm$. But $|G|/{p_1^{k_1}}$ is relatively prime to $p_1$, so $|G|/{p_1^{k_1}}$ divides $m$, and $mg_2=0$. Thus $mg_1=0$ as well, so also $p_1^{k_1}$ divides $m$ and thus $|G|$ divides $m$ and $g_1+g_2$ is a generator.

• Do you mean $g_1 \oplus g_2$ is a generator of $G$? I couldn't get how $mg_1 = -mg_2$ in that case? – MathMan Apr 20 '14 at 16:11
• No, $g_1+g_2$, using additive notation for the group operation in $G$. Then $m(g_1+g_2)=mg_1+mg_2=0$. – Kevin Carlson Apr 20 '14 at 17:32

Hint: decompose $G$ into the direct sum of its primary components; then examine each primary component and deduce it's cyclic because it has a minimum nonzero subgroup (that is, the intersection of its nonzero subgroups is nonzero).

• Can we be sure that $G$ can be decomposed into the direct sum of it's primary components? – MathMan Apr 20 '14 at 16:04
• @VHP It's a finite group, so it's torsion. – egreg Apr 20 '14 at 16:05
• Oh, I haven't studied about torsion in this book ( Gallian) as of yet. I will try looking it up though if i am able to understand anything at all :) – MathMan Apr 20 '14 at 16:07
• @VHP It's easy: consider $G[p]=\{x\in G:p^nx=0\text{ for some$n>0$}\}$; then $G[p]$ is a subgroup of $G$ and $G$ is the direct sum of the (nonzero) subgroups of the form $G[p]$, where $p$ runs through the prime numbers. Now, every group $G[p]$ is a $p$-group; if you show they are cyclic you're done, because the various components have coprime orders, so their direct sum is again cyclic. – egreg Apr 20 '14 at 16:09