Can u help me to prove this theorem,

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

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- Aug 10th 2008, 02:03 AMdimukFree groups
Can u help me to prove this theorem,

Two free groups are isomorphic if and only if they have the same rank. - Aug 10th 2008, 11:36 AMNonCommAlg
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 $\displaystyle R$ is a commutative ring with identity and $\displaystyle M$ a free $\displaystyle R$-module, then every two basis of $\displaystyle M$ have the same cardinality.

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

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

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

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

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

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

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

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