Free group generated by group

Question

Let $G$ be a group, let $F(G)$ be the free group generated by $G$.

Is it true that $F(G) \cong G$?

By the universal property of free groups there exists a unique group homomorphism $\eta: F(G) \rightarrow G$ such that $\eta \circ \phi=\text{Id}_G$ (here $\phi: G \rightarrow F(G)$ is the inclusion map of $G$ into $F(G)$).

It follows that $\eta$ is surjective, and there is left to show that it is injective (if this is the right path). To show it I thought of reasoning on the words of $F(G)$ as follow. Suppose $w=``g_1^{k_1}...g_N^{k_N}"$ is a word of $F(G)$ such that $\eta(w)=1_G$. Then $\eta(w)=\prod_{i=1}^N g_i^{k_i}=1_G$ by the properties of $\eta$.

Is it enough to conclude that $w$ must be $1_{F(G)}$? I think so because $\phi$ is not just an injective map here, it is also an injective group homomorphism, therefore this solve the problem.

Answer 1

For $G$ take the trivial group.

Then e.g. $\mathbb Z$ serves as free group generated by (singleton) $G$.

It is evident that $G$ and $\mathbb Z$ are not isomorphic.

Answer 2

Note that $F(X)\cong F(Y)$ when $X,Y$ are sets of the same cardinality. To be precise, any bijection $\rho\colon X\to Y$ induces an isomorphism $F(X)\to F(Y)$ by defining it on the generators using $\rho$ and extending using the property of group homomorphisms. This works precisely because $F(X)$ is a free group on the generators $X$.

Now if it was true that $G\cong F(G)$ for a group $G$, this would imply that for two groups $G$ and $H$ of the same cardinality we had $$ G\cong F(G) \cong F(H) \cong H. $$ But this must be false: for example the cyclic group $C_4$ and the product $C_2\times C_2$ both have $4$ elements but are not isomorphic.