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I was reading a solution of a linear algebra (or Euclid Geometry) problem which asks to

show that dihedral group $D_n$ of order $2n$ is isomorphic to the group $G=\left<a,b\right>$ s.t $a^2=e, \; b^2=e, \; (ab)^n=e$.

The solution first defines some map $\Phi:D_n \to G$ s.t $S \mapsto ab$ and $T \mapsto a$. (Here we denote $S$ as rotation about the origin for $2\pi / n$, and $T$ is some reflection in a line that goes through the origin.)

The solution then claims that "to show $\Phi$ is a well-defined homomorphism, it is sufficient to show that $\Phi (S^n)=\Phi (T^2) = \Phi(TSTS)=e \in G$"

$\Phi (S^n)=\Phi (T^2) = \Phi(TSTS)=e$ is really clear, but why can this be a sufficient condition of $\Phi$ being well-defined?

I wonder if it is due to the special character of the Dihedral Group, which is $T^2=S^n=TSTS=e$.

Or is it a general property of a well-defined homomorphism? (I mean if a "map" sends the identity to the other identity, then the "map" is well-defined.)

I don't think it's the general property. But, since I am not really familiar with the Group Theory, I would like to get some enlightenment.

Andreas Caranti
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john
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    This is a consequence of von Dyck's Theorem. If $G$ is a group given by generators $x_1,x_2,\ldots$, and relations $w_1,w_2,\ldots$, and $H$ is a group, then given elements $h_1,h_2,\ldots\in H$ there is a unqiue homomorphism $G\to H$ sending $x_i$ to $h_i$ if and only if the relations $w_1,\ldots,$ hold in $H$ with the $h_i$ in place of the $x_i$. – Arturo Magidin Feb 17 '22 at 18:55

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$\newcommand{\Set}[1]{\left\{ #1 \right\}}$First of all, what is really meant by your definition of $G$? This group $G$ is the quotient of the free group $F$ on the two-element set $\Set{a, b}$ by the smallest normal subgroup $N$ of $F$ containing $a^{2}, b^{2}, (a b)^{n}$.

By the universal property of free groups, there is a surjective homomorphism $\psi : F \to D_{n}$ mapping $a \mapsto T$ and $b \mapsto T S$. Clearly $N \le \ker(\psi)$.

To show that $N = \ker(\psi)$ it is enough to prove that $F/N$ has order at most $2 n$. To prove this, it is enough to show that every element of $F/N$ can be written in the form $b^{i} (a b)^{j} N$, for $0 \le i < 2$, $0 \le j < n$. The latter fact you prove using the fact that $a^{2}, b^{2}, (a b)^{n} \in N$.

Andreas Caranti
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  • Thank you very much! I just meant $G$ is the group generated by $a,b$ s.t order of $a,b$ are $2$. And, the other condition for this $G$ is that the order of $ab$ is $n$. – john Feb 18 '22 at 06:42