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The representation of the Quaternion group is $$\mathrm{Q} = \langle -1,i,j,k \mid (-1)^2 = 1, \;i^2 = j^2 = k^2 = ijk = -1 \rangle.$$

Does this imply that as long as I have found a group with $4$ generators, with one element "$-1$" that has order 2 and 3 elements $i,j,k$ that satisfy the equation $i^2 = j^2 = k^2 = ijk = -1$, I have proved that the group is the Quaternion group?

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  • $\begingroup$ What is your definition of "a quaternion group" ... ?? $\endgroup$ – DonAntonio Sep 27 '13 at 22:27
  • $\begingroup$ Hi @DonAntonio, I define $\mathrm{Q}$ as so ($\mathrm{Q} = \langle -1,i,j,k \mid (-1)^2 = 1, \;i^2 = j^2 = k^2 = ijk = -1 \rangle$). But I am not sure if it still applies to 8 elements from the fundamental group of CW complex. $\endgroup$ – Tumbleweed Sep 27 '13 at 22:36
  • $\begingroup$ Well, if you can spot one element of order two, three of order four and they fulfill the relations that you mention then they form a group isomorphic with the quaternions. $\endgroup$ – DonAntonio Sep 27 '13 at 22:38
  • $\begingroup$ Oh right, thanks. I wasn't certain about the equivalences(like $i^2 = -1$ implies $|i|=4$), thank you for confirming! $\endgroup$ – Tumbleweed Sep 27 '13 at 22:39
  • $\begingroup$ The fact that $i^2=j^2=k^2=ijk=-1$ uses all four generators, the last of which being the only one that is not over four and thus is two by supposition shows the development is $Q$. $\endgroup$ – Pax Kivimae Sep 27 '13 at 23:15
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You have some group $G$ you want to see is isomorphic to $Q$. You have 4 elements $a$, $b$, $c$, $d$ in $G$ that you think correspond to $i$, $j$, $k$, $-1$. So, form the candidate bijection $f: \{i,j,k,-1\} \to \{a,b,c,d\}$. By hypothesis, $f(i)$, $f(j)$, $f(k)$, $f(-1)$ satisfy the relations in the presentation of $Q$, so we know $f$ can be extended to a homomorphism $\phi: Q \to G$. Every proper subgroup of $Q$ contains $-1$, but $\phi(-1) \neq 1$. Therefore $\ker(\phi)=1$. Since you already knew $a$, $b$, $c$, $d$ generate $G$, you must have $\phi(Q) =G$. Therefore $\phi$ is an isomorphism and $G \cong Q$.

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  • $\begingroup$ I don't see why you could conclude the injectivity of $\phi$ so easily. Is there a general criterion you are using here? $\endgroup$ – Dune Sep 28 '13 at 13:58
  • $\begingroup$ I've added an argument to the answer. $\endgroup$ – TBrendle Sep 28 '13 at 21:20

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