# Is this a proof? (Of a simple result in abelian groups.)

In Topics in Algebra by Israel Nathan Herstein, there is a problem (2nd edition, page 35, number 4) that asks:

If $G$ is a group in which $(a \cdot b)^i = a^i \cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$, show that $G$ is abelian.

In other words, one can use, together with the usual rewrite rules for a group, the following three:

\begin{align} \tag{1} (a \cdot b)^i & = a^i \cdot b^i \\ \tag{2} (a \cdot b)^{i+1} & = a^{i+1} \cdot b^{i+1} \\ \tag{3} (a \cdot b)^{i+2} & = a^{i+2} \cdot b^{i+2} \end{align}

I will now put forward my attempt at a proof. My aim is to determine if I understand proof writing and whether I could compose a solid proof of a simple enough theorem.

Before departing towards the main goal, I saw it fit to establish the following two intermediate results:

Lemma 1: $\quad$ $b \cdot (a \cdot b)^n \cdot a = (b \cdot a)^{n+1}$

(I think the problem is marked with a star due to this part. It appears to me the creative core of the proof, as it requires something more than rewriting by rules.)

Proof: $\quad$ $(a \cdot b)^n$ is merely a shorthand notation for a sequence $a \cdot b \cdot \> \dots \> \cdot a \cdot b$ of appropriate length, witn $a$ and $b$ alternating. Prepending $b$ and appending $a$, we get $b \cdot a \cdot \> \dots \> \cdot b \cdot a$ — also an alternating sequence, evidently $= (b \cdot a)^{n+1}$. $\quad\square$

Lemma 2: $\quad$ $(a \cdot b)^i = (b \cdot a)^i$

Proof: $\quad$ If we show that $b \cdot (a \cdot b)^i \cdot a = b \cdot (b \cdot a)^i \cdot a$, we may then cancel equal prefixes and suffixes and thus reach the goal. So let us show: \begin{align} b \cdot (a \cdot b)^i \cdot a & = b \cdot (b \cdot a)^i \cdot a \\ (b \cdot a)^{i+1} & = \dots & \text{— Lemma 1.} \\ \dots & = b \cdot b^i \cdot a^i \cdot a & \text{— 1 left to right.} \\ \dots & = b^{i+1} \cdot a^{i+1} & \text{— Power notation.} \\ \dots & = (b \cdot a)^{i+1} & \text{— 2 right to left.} \\ \end{align} $\quad\square$

Now, let us return to the main goal.

Theorem: $\quad$ $b \cdot a = a \cdot b$

Proof: $\quad$

\tag{Part 1} \begin{align} (a \cdot b)^{i+2} & = a^{i+2} \cdot b^{i+2} & \text{— 3 per se.} \\ a \cdot b \cdot (a \cdot b)^i \cdot a \cdot b & = \dots & \text{— Power notation.} \\ \dots & = a \cdot a \cdot a^i \cdot b^i \cdot b \cdot b & \text{— Power notation.} \\ \dots & = a \cdot a \cdot (a \cdot b)^i \cdot b \cdot b & \text{— 1 right to left.} \\ b \cdot (a \cdot b)^i \cdot a & = a \cdot (a \cdot b)^i \cdot b & \text {— Cancel equal prefixes & suffixes.} \\ \end{align}

\tag{Part 2} \begin{align} (b \cdot a)^{i+1} & = \dots & \text{— Lemma 1.} \\ (b \cdot a)^i \cdot (b \cdot a) & = \dots & \text{— Power notation.} \\ \dots & = a \cdot a^i \cdot b^i \cdot b & \text{— 1 right to left.} \\ \dots & = a^{i+1} \cdot b^{i+1} & \text{— Power notation.} \\ \dots & = (a \cdot b)^{i+1} & \text{— 2 right to left.} \\ \dots & = (a \cdot b)^i \cdot (a \cdot b) & \text {— Power notation.} \\ \end{align}

Now, as, per lemma 2, $(a \cdot b)^i = (b \cdot a)^i$, we may cancel these prefixes. $\quad\square$

My questions:

• Is this a proof?
• Is it correct?
• Can it be improved?
• Proving lemma 2 you start from the claim and end up with an identity. That's a bit dangerous. You should start from something which is true (like the identity) and end up with the claim. Or start from the left side and end up with the right side of the claim. – Leppala May 31 '18 at 9:00
• @Leppala Are you saying that it is incorrect or that it is unnecessarily convoluted? – Ignat Insarov May 31 '18 at 10:06
• There are no mistakes as such. It's mostly about clear presentation but it's also easy to make logical mistakes when you write like that. You should at least make it clear whether you have equivalence or implication between the lines. – Leppala May 31 '18 at 10:17
• I think the proof of lemma 1 isn't fully rigourous (though it would be accepted by most) because "$(ab)^n$ is merely an abbreviation for..." is false : $(ab)^n$ is defined by induction. So I think a perfectly rigourous proof of lemma 1 should proceed by induction – Max May 31 '18 at 12:51
• @Max Oh, that is true. I did not think about it. I should have been suspicious but the power notation is so ubiquitous I let an assumption slip in. – Ignat Insarov May 31 '18 at 12:57

## 1 Answer

The proof isn't fully rigourous as you write something like $$x^n = x....x$$, which is how we think of $$x^n$$ and essentially how we use it, but it's not rigorous.

To make the intuition of $$x...x$$ precise, the main tool is induction (and things we proved before)

For instance, for lemma 1, here's one way to proceed by induction

For $$n=0$$, you are claiming $$b (ab)^0 a = (ba)^1$$. But $$(ab)^0 = 1$$ and $$(ba)^1 = ba$$ so by the fact that $$1$$ is neutral we are done. Assume the result is known for $$n$$ (the result being "for all $$a,b$$, $$b(ab)^n a = (ba)^{n+1}$$" : it is important to have "for all $$a,b$$" in the hypothesis !) and let's try to prove it for $$n+1$$. So let $$a,b$$ and compute : $$b(ab)^{n+1}a = b a (ba)^n b a$$ by using the induction hypothesis on $$(b,a)$$ (that's where we use " for all ").

Thus $$b(ab)^{n+1} a = (ba)(ba)^n (ba) = (ba)^{n+2}$$ (by some result we should prove earlier that $$x^nx^m = x^{n+m}$$, again by induction), and this is what we wanted for $$n+1$$.

By induction, we are done.