I am trying to self-learn group theory from an online pdf textbook, but some of the exercises are tough and the answer section doesn't help much (it only says "True"). 1. The problem statement, all variables and given/known data Prove or disprove: Every abelian group of order divisible by 3 contains a subgroup of order 3. 2. Relevant equations Not really sure. 3. The attempt at a solution I think I found a solution, but I am using a theorem which has not yet been proven in the textbook, it was only mentionned and said that they will prove it a few chapters later. Suppose G is a group of order 3n. The condition that G has a subgroup of order 3 is equivalent to the condition that G has an element of order 3 (because a group of order 3 must be cyclic by Lagrange's Theorem). But this condition is also equivalent to the condition that G has some element of order 3k, where k is 1, 2, ..., or n. (ie a multiple of 3). This is true because if there were such an element g, then g^k would have order 3 because the order of g^k is 3k / gcd( k, 3k ) = 3k/k = 3. And obviously the converse is true. So I have reduced the problem to that of proving that every abelian group G of order 3n has at least one element whose order is a multiple of 3. And this is where I got stuck. I had to use the Fundamental Theorem of Abelian groups, which is not yet proven. Assuming this theorem, we know G is isomorphic to the direct product of Z_i where the i are prime powers (including powers of 3 since G has 3n elements). Then consider an element of the form (0,0,...,x,0,...,0) where all entries are 0 except one corresponding to the group Z_3^p. The order of this element must divide 3^p, and is therefore a multiple of 3. By isomorphism, G also has an element of order divisible by 3, and the theorem is proven. Since my proof is assuming a theorem which was not proven yet, I feel it is incomplete. Is there are simpler way to answer the question without using the Fundamental Theorem?