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Fall 2016 - Problem 1

group theory

Let $G$ be a group generated by $a$ and $b$ with only relation $a^2 = b^2 = 1$ for the group identity $1$ . Determine the group structure of $G$ and justify your answer.

Solution.

$G$ is the free product $C_2 * C_2$ . Indeed, we have the maps $\func{\phi_1, \phi_2}{C_2}{G}$ given via $\phi_1\p{x} = a$ and $\phi_2\p{x} = b$ , which are well-defined since $a^2 = b^2 = 1$ . Now suppose $H$ is another group, and let $\func{\psi_1, \psi_2}{C_2}{H}$ be group homomorphisms.

Suppose we had a map $\func{f}{G}{H}$ such that the triangle commutes:

\p{f \circ \phi_i}\p{x} = \psi_i\p{x} \quad\text{for } i = 1, 2.

Since $\phi_1\p{x}, \phi_2\p{x}$ generate $G$ , this uniquely determines $f$ . Moreover, when defined this way, $f$ is a group homomorphism as

\p{f \circ \phi_i}\p{x}^2 = \psi_i\p{x}^2 = \psi_i\p{x^2} = 1

for $i = 1, 2$ . Hence, $G$ has the universal property of the free product $C_2 * C_2$ , so $G \simeq C_2 * C_2$ .