Was my abstract algebra textbook trying to kill me?

Solution 1:

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 $G$.

Solution 2:

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.