In teaching my algebraic topology class, this group showed up as part of an easy fundamental group computation: $\langle a,b\mid a^2=b^2\rangle$. My first instinct was that this must be $\mathbb{Z}*\mathbb{Z}/2$ because clearly every element can be written as a product of $b$'s (only to the power 1) and powers of $a$. But this turns out to be far from clear (and likely wrong). I assume this must be a well-known group to group theorists, so I'm curious if it's isomorphic to something that can be described by other means (or what's known about it in general).


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    $\begingroup$ This group is a free product of two copies of $\mathbb{Z}$ amalgamated along the subgroup $2 \mathbb{Z}$. See en.wikipedia.org/wiki/… and many books on infinite group theory (eg Rotman, or de la Harpe's book on geometric group theory, or Serre's book on Trees). $\endgroup$ Commented Dec 13, 2010 at 21:27
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    $\begingroup$ A quick proof that your suspicion was wrong: $\mathbb{Z}*\mathbb{Z}/2$ has torsion, whereas it's an easy exercise in Bass--Serre theory that an amalgamated product of two torsion-free groups is torsion-free. In particular your group, $\mathbb{Z}*_{2\mathbb{Z}}\mathbb{Z}$, is torsion-free. $\endgroup$
    – HJRW
    Commented Dec 13, 2010 at 22:37
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    $\begingroup$ @Greg Friedman: I think it would be generally useful if you could add in what computation the aforementioned group has shown up, just as a short background. $\endgroup$
    – M.G.
    Commented Dec 13, 2010 at 22:45
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    $\begingroup$ @Henry Wilton: One can also use the theory of one-relator groups to prove that this group is torsion free; for $r$ not a proper power in $F(X)$, $\langle X; r^n\rangle$ contains torsion if and only if $n > 1$ (see Magnus, Karrass and Solitar, or Lyndon and Schupp). $\endgroup$
    – ADL
    Commented Dec 14, 2010 at 8:17
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    $\begingroup$ Alan - of course, there are any number of ways that one can see it. It also follows from the observation that it's the fundamental group of the Klein bottle, for instance. In fact, perhaps that's the easiest argument. Using the theory of one-relator groups is probably one of the hardest. $\endgroup$
    – HJRW
    Commented Dec 14, 2010 at 21:35

2 Answers 2


Setting $c:=b^{-1}$ one obtains the presentation $$G= \langle a,c \, | \, a^2c^2=1 \rangle,$$ which is the fundamental group of the Klein bottle.

It is well known that another presentation of such a group is $$G= \langle x,y \,|\, x^{-1}yx=y^{-1} \rangle,$$ and this allows one to write $G$ as a semi-direct product of infinite cyclic groups, namely $$G= \mathbb{Z} \rtimes_{\sigma} \mathbb{Z}$$ where $\sigma \colon \mathbb{Z} \to \mathbb{Z}$ is defined by $\sigma(y)=y^{-1}$ and $\mathbb{Z}$ is written multiplicatively.

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    $\begingroup$ To make the "well known" other presentation explicit in terms of the original presentation, set x = aba^{-1} and y = ab^{-1} (conversely, a = xy and b = y^{-1}xy). $\endgroup$
    – KConrad
    Commented Dec 14, 2010 at 18:48
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    $\begingroup$ Even simpler: x=a, y=b^{-1}a. $\endgroup$
    – Leonardo
    Commented Nov 20, 2011 at 17:29

$a^2 = b^2$ says to me "two Mobius bands glued along their boundaries" which then says "the fundamental group of the Klein bottle." You can hear similar sounds from the presentations $\langle a, b \mid a^p = b^q \rangle$ which give rise to fundamental groups of torus knot complements. (Hmm. As long as $p, q$ are relatively prime.)


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