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In group category if $F_1$ be a free object on $X_1$ and $F_2$ is free object on $X_2$ and $F_1$ isomorphic to $F_2$ prove that |$X_1$|=|$X_2$|

whats the relationship between isomorphisms of free objects and cardinality of $X_i$ ?

Zhen Lin
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Rahmani
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    Hint: consider the abelianisations of $F_1$ and $F_2$ and use the fact that free abelian groups have a well-defined rank. – Zhen Lin Oct 27 '13 at 15:57

2 Answers2

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Consider the abelization $G^{ab}=G/{[G,G]}$, where $[G,G]$ is the commutator-subgroup of $G$.

If $F_1\cong F_2$, we get $$\mathbb{Z}^{|X_1|}\cong F_1^{ab}\cong F_2^{ab}\cong \mathbb{Z}^{|X_2|}.$$ Tensoring with $\mathbb{Q}$ over $\mathbb{Z}$, one gets $$\mathbb{Q}^{|X_1|}\cong \mathbb{Z}^{|X_1|}\otimes\mathbb{Q}\cong\mathbb{Z}^{|X_2|}\otimes\mathbb{Q}\cong\mathbb{Q}^{|X_2|}.$$ This means we have an isomorphism $\mathbb{Q}^{|X_1|}\cong\mathbb{Q}^{|X_2|}$ of $\mathbb{Q}$-vectorspaces and since the dimension of vectorspaces is well-defined even in infinity dimensions (dimension theorem) we get $|X_1|=|X_2|$.

archipelago
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An alternative proof is to observe that the cardinality of the underlying set of $F(X)$ equals the cardinality of $X$.

HeinrichD
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