Results 1 to 2 of 2

Math Help - Free groups

  1. #1
    Junior Member
    Joined
    Jul 2008
    Posts
    54

    Free groups

    Can u help me to prove this theorem,

    Two free groups are isomorphic if and only if they have the same rank.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by dimuk View Post
    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
    Last edited by NonCommAlg; August 10th 2008 at 01:52 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: September 29th 2010, 10:50 AM
  2. Quotient Groups - Infinite Groups, finite orders
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: August 11th 2010, 07:07 AM
  3. free groups, finitely generated groups
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 23rd 2009, 03:31 AM
  4. Free Groups
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 21st 2009, 04:16 AM
  5. Order of groups involving conjugates and abelian groups
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: February 5th 2009, 08:55 PM

Search Tags


/mathhelpforum @mathhelpforum