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For demonstration, let us consider the dihedral group $D_4$ which my instructor says has a group presentation $\langle a,b \mid a^4,b^2,abab \rangle$. Now, I understand that we can use van Dyck's theorem to show that there is an epimorphism $\frac{F}{N} \to D_4$ where $F$ is the free group on $\{a,b\}$ and $N$ is the normal subgroup generated by the relators. Therefore $|\frac{F}{N}|\geq |D_4|=8$.

The next step is to show that $|\frac{F}{N}|\leq 8$ so that we can conclude that the epimorphism is actually an isomorphism. I understand intuitively that we can just count $a^ib^jN$ for $i \in \{0,1,2,3\}, j\in\{0,1\}$. But how sure are we that these combinations of $a^{i}$ and $b^j$ generates non identical cosets when we (I, mostly) are not even sure of what are the elements in $N$? I know that the relators are in $N$, and is contained in the Kernel of the epimorphism above, but I'm still having a hard time understanding why the cosets above are not identical. Can someone enlighten me on this?

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If two of the cosets were identical then $|F/N|$ would be less than 8.

There are two parts to the argument. The first part is to prove that $|F/N| \ge 8$, and the second part is to prove that $|F/N| \le 8$. These two facts together imply that $|F/N| = 8$.

For the second part, we are not worried about repeated cosets. We know that $$F/N = \{N, aN, a^2N, a^3N, bN, abN, a^2bN, a^3bN\}$$ so $|F/N|$ is at most 8. Repetition would only decrease the number below 8; but we already know (from part 1) that $|F/N$| is at least 8. So repetition would create a conflict with the first part of the argument.

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  • $\begingroup$ Um, why? That's exactly what I'm asking. The books say "it's easy to show that the elements in $F/N$ are of the form $a^i b^j N$". But they don't really offer any explanation about why these cosets will not be identical to each other. $\endgroup$ – Kurome Sep 26 '16 at 4:05
  • $\begingroup$ What he seems to be saying is that it doesn't matter if they're identical. $\endgroup$ – mathematician Sep 26 '16 at 4:48
  • $\begingroup$ Yes. As far as the second part of the argument is concerned, we don't care if two of the cosets are identical, since we are proving "less than or equal". $\endgroup$ – Dave Radcliffe Sep 26 '16 at 5:02
  • $\begingroup$ Thanks, is this the standard strategy for showing presentations of finite groups? For example: $\langle a,b \mid a^4, b^3, abab \rangle$ will be the presentation for $S_4$ if I can show that some identifications of $a$ and $b$ can generate it and also satisfies the relator properties, then I proceed with a similar argument? $\endgroup$ – Kurome Sep 26 '16 at 6:33

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