Cyclic Groups in Artin's Algebra

Question

I'm currently working through Michael Artin's Algebra in my spare time, and I seem to be stuck on an easy question in section 2.4: Cyclic groups.

Let a and b be elements of a group G. Assume that a has order 7 and that $a^{3}b = ba^{3}$. Prove that $ab = ba$

I often find myself stuck with these reduction proofs.

What I know:

• If G is commutative then the proof is obvious. So let us only consider the case where G is not commutative.
• The element a has order 7, so a is a cyclic subgroup of G
• It follows from this that there are two cases:
  1. b is an element in the subgroup formed by a and can be represented as $a^{x}$ for some integer x
  2. b is not an element in the subgroup formed by a

Case 1:

We rewrite $$a^{3}b = ba^{3} => a^{3}a^{x} = a^{x}a^{3} => a^{3 + x} = a^{x + 3}$$ Multiplying both sides by $a^{-2}$ $$a^{3 + x - 2} = a^{x + 1} = ba$$ But without commutativity, $a^{3+x-2}$ doesn't reduce to $ab$, so I'm still stuck.

I can't seem to reduce this equation using inverses. My next thought was to try to use the fact that if an element say p has an inverse say q then both $pq = 1$ and $qp = 1$. Likewise with the inverse. This results in the same situation as above.

My intuition is that whatever the solution is, it is probably independent of whether or not b is an element of the subgroup formed by a.

I'm certainly making this problem much hard than it really is. Any hints would be greatly appreciated.

Answer 1

Hint: Try multiplying your equation $a^3b = ba^3$ by $ba^3$ on the left. This should reduce your equation to $a^{-1} b = ba^{-1}$ from which the result follows.

Answer 2

Hint: Use $a^7 = 1$ to put $a^7$ on the right side of $ab$, and then apply the relation $a^3b = ba^3$ multiple times to move $a$ to the other side.

Answer 3

Another possible approach to this question: since $a$ has order $7$ the subgroups generated by $a$ and $a^3$ are the same. From $a^3b = ba^3$ it follows that for every $k \in \mathbb Z$ the equality $a^{3k} b = b a^{3k}$ holds, since for some $k \in \mathbb Z$ we have $a^{3k}=a$ the thesis follows.