Simplified version of the following proof: If $G$ is generated by $\{a,b\}$ and $ab=ba$, then $G$ is abelian.

Question

I have seen a proof by induction for this theorem (located here: if G is generated by {a,b} and ab=ba, then prove G is Abelian), but I wondered if my version had any validity...or if it is too "informal".

The problem statement is as follows:

Suppose a group $G$ is generated by two elements $a$ and $b$. If $ab=ba$, prove that G is abelian.

Firstly, if $ab=ba$, then you know an element of the form $a^n \circ b^m$ can be "shuffled" into any permutation. For example, if $x=a \circ b \circ b \circ b$, then because of the property of $ab=ba$, we know that $x$ can be rewritten in 3 additional manners:

$x=b \circ a \circ b \circ b$

$x=b \circ b \circ a \circ b$

$x=b \circ b \circ b \circ a$

Therefore, if we also know that $a,b$ generate all elements within $G$, it should be valid to claim that, for an arbitrary element $x$, $x=a^n \circ b^m$...because these elements can be shuffled into all possible composition arrangements.

Moving forward, suppose $x,y \in G$ where $y=a^i \circ b^j$ and $x=a^m\circ b^n$

I must show that $xy=yx$:

Beginning:

$xy=(a^m \circ b^n)\circ (a^i \circ b^j)$

$=a^m \circ (b^n \circ a^i) \circ b^j$

$=a^m \circ (a^i \circ b^n) \circ b^j$

By simple "regrouping" of common $a$ terms and $b$ terms

$=a^i \circ (a^m \circ b^j) \circ b^n$

$=(a^i \circ b^j)\circ(a^m\circ b^n)$

$=yx$

Is this too informal of a proof? (i.e. should I stick with the induction method?)

Answer 1

You have explicitly shown any words of the form $a^m b^n$ can be arbitrarily permuted (although arguing by example is not great). You have explicitly shown that words of the form $a^m b^n a^i b^j$ are equal to $a^i b^j a^m b^n$. Things that are missing:

• An argument for words having more than four "islands" of repetitions of the same term.
• Maybe: anything involving inverse elements. You don't say what set from which you draw $m$, $n$, $i$, or $j$, so I don't know if they can be negative.
• Maybe: any word that doesn't start with an $a$. You don't really constrain $m$, $n$, $i$, or $j$, so I don't know if any of them can be zero.
• $a^m b^n a^i b^j = a^{m+i}b^{n+j}$. Which, ultimately is what you really want: each word in $a$ and $b$ are in an equivalence class under permutation and there is a canonical choice of representative from the class.

Some of this can be patched into your argument. But the inductive argument should handle all of this without so much patching.

Answer 2

That's the general idea of the result, yes, so, as such, it could be construed as a proof, albeit, indeed, a somewhat informal one.

Answer 3

Let $H$ be the set of all elements of $G$ of the form $a^ib^j$.

Now consider any two elements of $H$, $a^rb^s$ and $a^tb^u$.

Then $a^rb^sa^tb^u=a^{r+t}b^{s+u}$ since $a$ and $b$ commute.

Then $H$ is closed and contains both $a$ and $b$. $H$ is therefore all of $G$.

Furthermore, we similarly have $a^tb^ua^rb^s=a^{r+t}b^{s+u}$ and so $G$ is abelian.

Answer 4

Here's a formal and induction-less proof: As $aa=aa$ and $ab=ba$, the Centralizer of $a$ contains $a$ and $b$, hence equals $G$. It follows that the centre of $G$ contains $a$. By the same argument, the centre contains $b$. As the centre contains $a $ and $b$, it must equal all of $G$, i.e., $G$ is abelian.