Free groups

Quote:

Originally Posted by dimuk

Can u help me to prove this theorem,

Two free groups are isomorphic if and only if they have the same rank.

if they have the same rank, then using the definition of a free group it's obvious that they must be isomorphic. the converse is not trivial. to prove it we'll first need a nice

lemma that extends a well-known fact about vector spaces to free modules over commutative rings with identity:

Lemma: (IBN for commutative rings) if $R$ is a commutative ring with identity and $M$ a free $R$ -module, then every two basis of $M$ have the same cardinality.

Proof: let $m$ be a fixed maximal ideal of $R$ and $X$ any basis for $M.$ clearly $M/mM$ is a vector space over $R/m.$ let $\bar{X}=\{x+mM: \ x \in X \}.$ if $x_1 + mM = x_2 + mM,$

for some $x_1 \neq x_2$ in $X,$ then $x_1-x_2=\sum a_jx_j$ for some $a_j \in m$ and $x_j \in X,$ which is impossible since $X$ is basis for $M$ and $1 \notin m,$ thus $|X|=|\bar{X}|.$ a similar argument

shows that $\bar{X}$ is a basis for $R/m$ -vector space $M/mM.$ thus $|X|=\dim _{R/m}M/mM,$ which completes the proof because $\dim_{R/m}M/mM$ does not depend on $X. \ \ \square$

Corollary: let $R$ be a commutative ring with identity and $M$ and $N$ be two free $R$ -modules with basis $X$ and $Y$ respectively. if $M \simeq N,$ then $|X|=|Y|.$

Proof: let $f:M \rightarrow N$ be an isomorphism. then $Y$ and $f(X)$ both are basis for $N,$ thus by the Lemma $|X|=|f(X)|=|Y|. \ \ \ \square$

now suppose $G_1$ and $G_2$ are two free groups on $X$ and $Y$ respectively. suppose $G_1 \simeq G_2.$ let $G_1'$ and $G_2'$ be their commutator subgroups. then clearly $G_1/G_1' \simeq G_2/G_2'.$

it's easy to see, from the definition of free groups, that $G_1/G_1'$ and $G_2/G_2'$ are free (obviously abelian) on $X$ and $Y$ respectively. so $G_1/G_1'$ and $G_2/G_2'$ can be considered

as two isomorphic $\mathbb{Z}$ -modules. the above corollary now shows that $|X|=|Y|. \ \ \ \square$