# Equivalent identification to get the projective plane?

I think $$\langle a,b | abab= 1 \rangle = \langle a,b | abba = 1 \rangle$$ are 2 equivalent presentations of the fundamental group of the projective plane. To show this, I have tried to transform one into the other using Tietze transformations and substituting $c = ab$. But it doesn't seem to lead anywhere.

Is what I want to show wrong? And if not: what tools apart from Tietze transformations are there?

If two groups are isomorphic then there is a Tietze transformation transforming one into the other so my failed attempt was actually supposed to work if the claim is right.

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Surely the (group-theory) tag wouldn't go amiss here...? – user1729 Jul 8 '11 at 14:01
By "the projective plane", do you mean the real projective plane $\mathbb{RP}^2$? If so, I'm pretty sure neither of those presentations is right, since the fundamental group is $\mathbb{Z}\mid 2\mathbb{Z}$ and those groups clearly have elements of infinite order. You may have gotten confused because the standard way to construct $\mathbb{RP}^2$ by identifying opposite sides of a square gives you more than one basepoint (unlike with a torus or Klein bottle). – MartianInvader Jul 8 '11 at 16:10

These groups are not equivalent. To see this, you can use the theory of one-relator groups. Specifically, a group with a single defining relator contains torsion if and only if the relator is a proper power (Thm 4.12 of Magnus, Karrass and Solitar's book "Combinatorial Group Theory"). That is, supposing $R$ is not a proper power in $F(X)$ then $\langle X; R^n\rangle$ has torsion if and only if $n>1$. Moreover, every element of finite order has order dividing $n$.

Now, $abab=(ab)^2$ so your first group contains torsion. Indeed, it contains elements of order $2$, but of order no more than $2$.

On the other hand, there does not exist $w\in F(a, b)$ such that $w^2=aabb$. This is by a paper of R. Lyndon, "The equation $a^2b^2=c^2$ in free groups", 1959, where he proves that if $w^2=x^2y^2$ in a free group then $[x, y]=1$. Clearly $[a, b]\neq 1$ in $F(a, b)$.

Therefore, one of your groups contains elements of order two while the other does not.

Indeed, there is a paper, "The isomorphism problem for two-generator one-relator groups with torsion is solvable" by Steve Pride (1977) which solves the isomorphism problem for (surprisingly!) two-generator one-relator groups with torsion. It says, $\langle a, b; R^n\rangle\cong \langle a, b; S^m\rangle \Leftrightarrow m=n$ and there exists $\phi \in \operatorname{Aut}(F_2)$ such that $R\phi=S$ (and one can decide if there exists such a $\phi$ by Theorem N2 of Magnus, Karrass and Solitar).

EDIT: Actually, it is perhaps easier just to note that one group is the free product of the infinite cyclic group with the cyclic group of order two, $\mathbb{Z}\ast C_2$, while the other is the fundamental group of the Klein bottle, which is well-known to be torsion-free (and as this is an algebraic topology question, proving that the fundamental group of the Klein bottle is torsion-free is an exercise left to the reader!).

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