# Abelian group of order 99 has a subgroup of order 9

Prove that an abelian group $G$ of order 99 has a subgroup of order 9.

I have to prove this, without using Cauchy theorem. I know every basic fact about the order of a group.

I've distinguished two cases :

• if $G$ is cyclic, since $\mathbb Z/99\mathbb Z$ has an element of order $9$, the problem is solved.

• if $G$ isn't cyclic, every element of $G$ has order $1,3,9,11,33$. I guess I need to prove the existence of an element of order $9$. How should I do that ?

Note that $G$ is abelian (I haven't used it yet).

Context: This was asked at an undergraduate oral exam where advanced theorems (1 and 2) are not allowed.

• By Cauchy's, you mean Lagrange's theorem? Dec 21, 2014 at 19:14
• @AaronMaroja No, I referer to this en.wikipedia.org/wiki/Cauchy%27s_theorem_%28group_theory%29 Dec 21, 2014 at 19:15
• The group need not have an element of order $9$ - the subgroup could be a product of two cyclic groups of order $3$. Dec 21, 2014 at 19:16
• @Krish - to clarify. I was commenting on the OP's statement "I guess I need to prove the existence of an element of order 9". There can be a subgroup of order $9$ without there being an element of order $9$ Dec 21, 2014 at 19:21
• Are you able to show that $G$ must have an element of order 3?
– user169852
Dec 21, 2014 at 19:24

There exists an element $a$ of order $3$. If there was not every element would need to have order $11$.

But then a counting argument leads to contradiction.

We take the quotient group $G/<a>$. Then it has order $33$. The same argument shows that there exists an element $b$ of order $3$. Take the map $p: G \rightarrow G /<a>$ the natural projection. The pre-image of a subgroup is a subgroup, so we take the subgroup of $G$ $p^{-1}(<b>)$. And this subgroup has order $9=|\rm{Ker p}|| <b>|$.

If you can show that $G$ has an element $g$ of order 3, then $\langle g \rangle$ is a normal subgroup (since $G$ is abelian), and therefore $G / \langle g\rangle$ is an abelian group. Since the order of $G / \langle g\rangle$ is 33, this group also contains an element, say $x + \langle g\rangle$, of order 3. If $n$ is the order of $x$ in $G$, then $(x + \langle g\rangle)^n = e$, so $n$ is divisible by 3. Therefore, $y = x^{n/3}$ is an element of $G$ with order $3$. Now you can check that the subgroup of $G$ generated by $g$ and $y$ has order 9.

So the problem reduces to showing that $G$ (or any finite group with order divisible by 3) has an element of order 3. This is proved quite similarly to the argument above, and essentially amounts to proving Cauchy's theorem for the special case of an abelian group. This is done adequately by Wikipedia, so I won't reproduce it here.

Direct counting of elements can handle this without even using factor groups (although to understand group theory better, it's good to understand the answers here based on factor groups).

There is one element of order $1$. This leaves $98$ elements to consider.

If even a single one is order $9$, it generates a subgroup of order $9$.

If any one of them has order $99$, then the group is isomorphic to $C_{99}=\langle g\rangle$, and so $\{1,g^{11},\ldots,g^{88}\}$ is a subgroup of order $9$.

Otherwise, assuming no elements of order $9$ or $99$:

• Each element of order $33$ can be grouped with $\varphi(33)=20$ other elements of order $33$, taking powers relatively prime to $33$.
• Each element of order $11$ can be grouped with $\varphi(11)=10$ other elements of order $11$, taking powers relatively prime to $11$.

Taking elements of order $33$ and $11$ together, we would only have some multiple of $10$. Therefore at a minimum there are $8$ elements having order $3$. Take any two of these where neither equals the other squared, and since the group is abelian they generate a subgroup isomorphic to $C_3\times C_3$.

Note that $99=3^2\cdot 11$ so any $3$-Sylow has order 9. By the Sylow theorems, the number of $3$-Sylow subgroups, $n_3$, satisfy $$n_3\equiv 1\mod 3\qquad\mbox{and}\qquad n_3|11$$ Thus, there is one $3$-Sylow subgroup (and it necessarily has order 9).

• The OP specifically said without Cauchy, which also excludes Sylow. Dec 21, 2014 at 19:44

Let's $A=\{e,a,a^2\}$ and $B=\{e,b,b^2\}$ are cyclic subgroups of order $3$ such that $A\cap B={e}$, then $H=AB$ has order $9$, otherwise we would have $a^ib^j=a^mb^n$ for some $i,j,m,n$, which is contradicts that $A\cap B={e}$. Now we only left to prove that there exist such $A,B$. But it follows from the fact that there can be only one subgroup of order $11$.

• Why can there only be one subgroup of order 11? Dec 21, 2014 at 21:24
• If there are two different subgroups their intersection is only ${e}$, because every element is generator. Now if we multiply them we will get that $a^ib^j=a^mb^n$, where $a,b$ are generators of these groups and $i\ne m$ or $j\ne n$,which is contradiction that their intersection is only $e$. Dec 21, 2014 at 21:43
• You seem to have a quite large leap in the logic there, since there are indeed abelian groups with more than one subgroup of order 11. Dec 21, 2014 at 21:46
• Yes, it is possible, but their size can't be less than $121$. Dec 21, 2014 at 21:51
• But your argument does not use that at all. Dec 21, 2014 at 21:53

If it's abelian then you know it's a product of cyclic groups. You know that the order of these cyclic groups can be $3, 9, 11, 33$, or $99$ and their product must be $99$. You've already seen that if $99$ is in their that you're done, and if $9$ is in their then you're done. So your left with a product of cyclic subgroups of orders $3, 11$, or $33$ and their product must be $99$.

So show that $\mathbb Z/33$ has a subgroup of order $3$ and you must have either this group and $\mathbb Z/3$ or you must have two copies of $\mathbb Z/3$. In either case this gives you a subgroup of order $9$.

• If you can't use Cauchy's theorem, I doubt you can use the structure theorem either! Dec 21, 2014 at 19:26
• This presumes that the OP is able to use the fundamental theorem of finite abelian groups, which might be a stretch if not even Cauchy's theorem is available.
– user169852
Dec 21, 2014 at 19:27
• Thanks for your answer, but I'm not supposed to use this theorem either... Dec 21, 2014 at 19:27
• Well, there are certainly ways to prove this that don't use the classification, clark just posted one for example, but they all basically amount to proving Cauchy's theorem for the special case of $p = 3$.
– Jim
Dec 21, 2014 at 19:38