Newbie question: generating the group elements from a group presentations

Question

I am reading about group presentation and I don't understand how to generate the group elements from the presentation when there are several relations.

Take the Klein $4$ group, which, according to Wikipedia, has 3 relations.

$$\langle a, b \mid aa, bb, abab \rangle.$$

In my mind that means that applying the two generators and replacing the relations should yield the four elements of Klein 4:

• $e$,
• $a$,
• $b$,
• $ab$

But I got three more elements:

• $aba$
• $bab$
• $ba$

What am I missing?

Answer 1

Lemma: For any $x$ in a group, $$x^2=e\iff x^{-1}=x.$$

Proof: Suppose $x^2=e$. Then

$$\begin{align} x^{-1}&=ex^{-1}\\ &=x^2x^{-1}\\ &=x(xx^{-1})\\ &=xe\\ &=x. \end{align}$$

Now suppose $x^{-1}=x$. Then

$$\begin{align} e&=x^{-1}x\\ &=xx\\ &=x^2.\,\square \end{align}$$

Now:

• $aba=b^{-1}=b$ by $abab$ and $b^2$,
• $bab=a^{-1}=a$ by $abab$ and $a^2$, and
• $ba=a^{-1}b^{-1}=ab$ by all three relators.

Answer 2

You must know the data that the relations are giving you. Typically, when relations are written out that way, it is implicit that all those terms are set equal to the identity $e$ in your group $G$.

In your case, the relation $aa$ really means $aa=e$. Multiplying by $a^{-1}$ gives $a=a^{-1}$. Likewise, we learn that $b=b^{-1}$.

From $abab=e$ we multiply on the right by $b^{-1}$ to learn $aba=b^{-1}$. Putting this all together, we see that $aba=b^{-1}=b$, which is NOT a new element.

You can handle the rest....