Let \(G_n\) be the free group on \(n\) generators. Show that \(G_2\) and \(G_3\) are not isomorphic.
Solution
Let \(C_2\) denote the cyclic group of order \(2\).
1. For every positive integer \(n\),\[ \abs{\Hom(G_n,C_2)}=2^n. \]
Fix a free basis \(x_1,\ldots,x_n\) of \(G_n\). By the universal property of the free group, a homomorphism \[ \varphi:G_n\to C_2 \] is determined uniquely by the freely chosen values \[ \varphi(x_1),\ldots,\varphi(x_n)\in C_2. \] Each generator has two possible images, independently of the others, so there are exactly \(2^n\) homomorphisms.
2. Isomorphic groups have the same number of homomorphisms to \(C_2\).
If \[ \psi:H\xrightarrow{\sim}K \] is an isomorphism, then precomposition with \(\psi\) gives a bijection \[ \Hom(K,C_2)\longrightarrow\Hom(H,C_2), \qquad \alpha\longmapsto\alpha\circ\psi. \] Its inverse is precomposition with \(\psi^{-1}\).
3. The groups \(G_2\) and \(G_3\) are not isomorphic.
By step 1, \[ \abs{\Hom(G_2,C_2)}=2^2=4, \qquad \abs{\Hom(G_3,C_2)}=2^3=8. \] If \(G_2\cong G_3\), step 2 would force these cardinalities to be equal, a contradiction. Thus \[ \boxed{G_2\not\cong G_3}. \]
4. Q.E.D.
Step 3 proves the required non-isomorphism.