Was my abstract algebra textbook trying to kill me? I'm doing an independent study (self-taught course) in abstract algebra and I'm using the abstract algebra textbook here: http://abstract.ups.edu/. In Chapter 9: Isomorphisms, problem 20 asks: "Prove or disprove: Every abelian group of order divisible by $3$ contains a subgroup of order $3$." I spent a lot of time on this question and eventually came up with the answer below, but this question seemed a lot harder than all of the other questions. Keeping in mind that the book hasn't yet given me all of the usual tools to prove this (see below or the textbook for the ones I do have), did I miss something obvious? Or is this actually that hard? At one point, the book says, "In Chapter 13, we will prove that all finite abelian groups are isomorphic to direct products of the form $\mathbb{Z}_{p_1^{e_1}}\times\cdots\times\mathbb{Z}_{p_k^{e_k}}$, where $p_1,\dots,p_k$ are (not necessarily distinct) primes." Do you think I was supposed to use this, even though the book hasn't proven it yet?
So far, I've covered some basic material about cyclic groups, the groups $S_n$, $A_n$, and $D_n$, Lagrange's Theorem, basic properties of isomorphisms, and some basic direct product stuff. In particular, I have not covered quotients or group actions. For more complete information, please see the textbook.
My solution (abbreviated)
Call a group $3$-free if its order is not divisible by 3. Let $G$ be a group with order divisible by $3$, and find a maximal $3$-free subgroup $H$ (one that is not contained in any other $3$-free subgroup). Pick a $g\in G\setminus H$ and let $d$ be the least positive integer such that $g^d\in H$. Then $g^iH=g^jH$ iff $i\equiv j\text{ mod }d$, so the subgroup (of $G$) $H'=\langle g\rangle H$ has $|H'|=d\cdot |H|$. $|H'|$ must be divisible by $3$, so $d$ is divisible by $3$, and therefore the order $k$ of $g$ is divisible by $3$, so $g^{k/3}$ has order $3$.
Thanks!
 A: I assume that Sylow theorems were not covered yet?
And since the theorem holds also for non-abelian groups, I wonder why they restrict to abelian groups. It makes the following easier though:
Let $G$ be a group of order $n$ divisible by $3$. Select $h\in G\setminus\{1\}$. If the order of $h$ is divisible by $3$, then you find a power of $h$ that has order $3$. Otherwise $G/\langle h\rangle$ has order divisible by $3$ and can be assumed by induction to have an element $g+\langle h\rangle$ of order $3$, which has order a multiple of $3$ in $G$.
A: Here's something using the idea from this proof of Sylow's theorem.
I doubt this is what the author of the book intended, but this solution does not use quotients or homomorphisms. All you need is the concept of maximal subgroup and the formula for the order of the product of two subgroups.
We can proceed by induction on the order $|G|$. The case $|G| = 1$ is vacuously true, so let $|G| > 1$ and assume the claim true for all Abelian groups of order $< |G|$. Since $G$ is finite and has order $> 1$, we can find a maximal subgroup $M < G$. Let $x \not\in M$ be an element of $G$. Because $G$ is Abelian, the product $M\langle x \rangle$ is a subgroup and thus $G = M\langle x \rangle$ by maximality of $M$. Now $3$ divides the order $|G| = |M \langle x \rangle|$. Hence by the formula for order of the product of subgroups, it divides $|M||\langle x \rangle| = |M||x|$. Since $3$ is a prime, it divides $|M|$ or $|x|$. If it divides $|M|$, the claim follows by induction. Otherwise it follows from the cyclic case which is easy.
Note that the only property of $3$ we used here is the fact that it is a prime. The same proof works if we replace $3$ by any prime.
