# $\mathbb Z_{p^n}$ is not the direct product of any family of its proper subgroups

I'm trying to solve this question of Hungerford's Algebra book:

$$S_3$$ is not the direct product of any family of its proper subgroups. The same is true of $$\mathbb Z_{p^n}$$.

The first claim is easy: we note that every subgroup of $$S_3$$ is of order $$2$$ or $$3$$ by Lagrange, since its subgroups are of prime order, they are cyclic, then abelian, but $$S_3$$ is not abelian, contradiction because $$S_3$$ is not abelian while the direct product of abelian groups has to be abelian.

My problem is with the second claim, I need help.

• $\Bbb{Z}_p^n$ is obviously a direct product. Perhaps you mean $\Bbb{Z}_{p^n}$? – Chris Eagle May 6 '13 at 17:47
• @ChrisEagle yes, of course, thanks I will edit the question. – user42912 May 6 '13 at 17:49
• Note that in your argument for the first case: the direct product of cyclic groups is not necessarily cyclic. In this case, indeed $\mathbb Z_2 \times \mathbb Z_3 \cong \mathbb Z_6$ and is necessarily cyclic, but that's because $\gcd(2, 3) = 1$. But $\mathbb Z_2$ is cyclic, however $\mathbb Z_2 \times \mathbb Z_2$ is not. – Namaste May 6 '13 at 18:04
• @amWhy I proved in the following way: $a_1,a_2\in A$ and $b_1, b_2 \in B$, $A, B$ cyclic groups, then $(a_1,b_1)\cdot (a_2,b_2)=(a_1a_2,b_1b_2)=(a_2a_1,b_2b_1)=(a_2,b_2)(a_1,b_1)$ what is wrong with my proof? thank you for your remark. – user42912 May 7 '13 at 1:39
• @amWhy yes, I understood now, I was confused with the terms abelian and cyclic, thank you again – user42912 May 7 '13 at 1:42

Hint:

$\Bbb{Z}_{p^n}$ has the single minimal subgroup.

• What is a single minimal subgroup? thank you for your answer. – user42912 May 6 '13 at 17:54
• Let $\Bbb{Z}_{p^n}$ is generated by $a$. Then this subgroup is generated by $a^{p^{n-1}}$. – Boris Novikov May 6 '13 at 18:04

Well, $$C_{p^n}$$ is isomorphic to the direct product of itself with the trivial group, but you probably want to exclude that case.

So suppose that $$C_{p^n}\simeq G\times H$$ for non-trivial groups $$G$$ and $$H$$. What are the possible orders of $$G$$ and $$H$$? Could $$G\times H$$ contain an element of order $$p^n$$?

• No it cannot! But then what is special about the group being of mod $p^k$? Could this be generalized to any cyclic group of order $n$, and not just prime to the k power? – Sally G May 2 '19 at 2:56
• No, the statement is not true for general $n$. It might be helpful to review the Chinese remainder theorem. Prime powers are special in that they cannot be decomposed into a nontrivial product of coprime integers. – carmichael561 May 2 '19 at 3:06
• No. Consider $C_6$. It really is the case that $C_6 \cong C_3 \times C_2$. To see this, consider the map $f : C_6 \to C_3 \times C_2$ given by $f(a) = (a, a)$. – Charles Hudgins May 2 '19 at 3:07
• For the record, this was an answer to a different question that was merged with this one. – carmichael561 May 10 '19 at 1:09

Hint

If ${\mathbb Z}_{p^n}$ is the direct product of two smaller subgroups $H_1,H_2$, then the order of any element in ${\mathbb Z}_{p^n}$ would divide $lcm(|H_1|, |H_2|)$. Since both of those orders are power of $p$, $lcm(|H_1|, |H_2|)=\max\{ |H_1|, |H_2| \}< p^n$ which contradicts the fact that ${\mathbb Z}_{p^n}$ is cyclic.

• Why the order of any element of $\mathbb Z_{p^n}$ divide $lcm(|H_1|,|H_2|)$? thank you for your answer. – user42912 May 6 '13 at 18:03
• @user42912 Exercise: in any abelian group, the order of $ab$ is the lcm of the orders of $a$ and $b$. – anon May 6 '13 at 18:12
• Because if $(a,b) \in H_1 \times H_2$ you have $a^{|H_1|}=e$ and $b^{|H_2|}=e$ thus $(a,b)^{lcm(|H_1|, |H_2|)}=(e,e)$. – N. S. May 6 '13 at 18:13
• @anon I've already prove that there is an element or order of the lcm of the orders of $a$ and $b$, but I didn't know this stronger result. – user42912 May 6 '13 at 18:19

Suppose you could write $\mathbb{Z}_{p^n} = A \oplus B$. Then you have $|A| = p^a$ and $|B| = p^b$ with $n = a+b$ and $a,b>0$.

$\mathbb{Z}_{p^n}$ has an element of order $p^n$, one can show that $A \oplus B$ does not.

Well, $${\Bbb Z}_{p^n}$$ has a unique maximal subgroup, but a direct product of non-trivial groups has not.

Can you prove that a finite cyclic group of order $n$ has precisely one subgroup of order $d$ for each divisor $d$ of $n$?

Hint. $$\sum_{d\mid n} \varphi(d) = n.$$

Now, what happens when we have a direct product of two groups of $p$-power order?

Hint:

1. Each subgroup of a $p$-group is a $p$-group.
2. Any nontrivial product of nontrivial $p$-groups is not cyclic.

The first point is trivial. For the second, notice that the order of a tuple is the least common multiple of orders of its coordinates.

In fact, you can use the same argument to show that a product of groups is cyclic iff each is cyclic and they have coprime orders.

• This is the way I thought of the problem, so naturally I like it. I might add one more hint involved in the proof of 2.: because subgroups of cyclic groups are cyclic, one immediately reduces to the case $Z_p \times Z_p$. – Pete L. Clark May 7 '13 at 1:55

If $\mathbb{Z}_{p^n}=H \times K$, then $H \cap K= \{0\}$. Suppose there exist $h \in H\backslash\{0\}$ and $k \in K \backslash \{0\}$. Then, you can find $u,v \in \mathbb{Z}$ such that $uh=vk \neq 0$ contradicting $H \cap K\{0\}$.

• Why $H\cap K=\{0\}$? thank you for your answer – user42912 May 6 '13 at 18:23
• You can see an element of $H$ (resp. $K$) as $(h,0)$ (resp. $(0,k)$). If $(a,b) \in K \cap H$, then both $a$ and $b$ have to be zero. – Seirios May 6 '13 at 20:35