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Consider the group $G = \langle a, b : a^4 = 1, a^2 = b^2, bab^{-1} = a^{-1}\rangle$. This is a standard presentation of the quaternion group. But why could it not be the Klein four group?

My professor told us that we should consider this notation to mean "$G$ is the largest group that satisfies such properties", and that would solve the problem, since one can exhibit the matrix form of quaternions and also show that $|G| \leq 8$. How does this relate to the definition of presentation as a quotient of a free group by a normal subgroup?

This is a follow-up question to Counting order of quaternion group defined by relations, motivated by professor Derek Holt's insightful comment.

hdecristo
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    The group is the "most free" group that satisfies the conditions. – Lucenaposition Jun 23 '25 at 23:09
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    That this group is not the Klein four is directly implied by the linked question: the klein four has order 4, not order 8. It seems like you already know this, so it's a bit unclear exactly what question you want answered. – Noah Caplinger Jun 24 '25 at 00:05

2 Answers2

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The Klein $4$-group is generated by $2$ elements $a$ and $b$ satisfying the relations you put on them when defining $G$, but that does not mean the group $G$ can be the Klein $4$-group. After all, the trivial group also has $2$ generators $a$ and $b$ satisfying the relations you put on them when defining $G$ but you would not have argued that $G$ is the trivial group, right?

The fact that multiple nonisomorphic groups can have generators satisfying the same relations in a group presentation is not an inconsistency: all those groups will be quotients of the biggest such group, by which I mean the group actually defined by the group presentation. By definition, a group presentation is a group of the form $F/N$ where $F$ is a free group with generators corresponding to each of the generating elements in the group presentation and $N$ is the smallest normal subgroup of $F$ containing the relations.

Example. The dihedral group $D_n$ of order $2n$ can be described as $\langle x,y|x^n = 1, y^2 = 1, yxy^{-1} = x^{-1}\rangle$, which is $F_2/N$ where $F_2$ is the free group on two elements $x$ and $y$ and $N$ is the normal subgroup of $F_2$ generated by $\{x^n, y^2, yxy^{-1}x\}$. Every quotient group of $D_n$ also has two generators satisfying the same relations as in the definition of $D_n$, but proper quotient groups of $D_n$ will have "more relations".

It is not always easy to tell how big a group is when it is described by a group presentation: some group presentations can be the trivial group and that can be very tricky to show. See here.

KCd
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Let $F_2 \cong \langle a,b \rangle$ be the free group on two generators. The presentation can be used to define a surjective homomorphism $\varphi : F_2 \longrightarrow Q_8$. Then, $F_2/\ker \varphi \cong Q_8$. Let $H$ be the smallest normal subgroup of $F_2$ that contains $\langle a^4, a^2b^{-2}, bab^{-1}a \rangle$. Since $H \le \ker \varphi$, we have a surjective homomorphism $F_2/H \longrightarrow F_2/\ker \varphi$. But, by definition, $G \cong F_2/H$, so $Q_8$ is a quotient of $G$. Together with a proof that $|G| \le 8$ (of which you seem to be already satisfied), it follows that $G \cong Q_8$.

The argument showing that $Q_8$ is a quotient of $G$ can be adapted to show that any group satisfying a presentation is a quotient of the group defined by that presentation.

Eric M. Schmidt
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