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I'm looking for a tangible example of a free abelian group whose quotient with a subgroup is not free abelian. There's a theorem that says that every abelian group is a quotient of some free group, but I'm looking for a more exact example.

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Isn't it obvious what the quotients of $\mathbb{Z}$ are? –  hardmath Dec 7 '12 at 2:56
    
A relevant reference: en.wikipedia.org/wiki/Free_abelian_group –  Jonas Meyer Dec 7 '12 at 3:19

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up vote 3 down vote accepted

The most obvious example of a free group is $\Bbb Z$--which is the free group of one generator. (Can you see why?)

Now consider any finite cyclic group--necessarily abelian but not a free abelian group, and (isomorphic with) a quotient of $\Bbb Z$.

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"The only non-trivial free abelian group" can be, imfho, a misleading phrase. Perhaps it'd be better to write "the only non-trivial free group which is abelian"...perhaps. –  DonAntonio Dec 7 '12 at 4:25
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Ah! My apologies--that's how I was interpreting it, but following Jonas's link from the comment above, I see that I was mistaken. I'll edit to prevent such confusion. –  Cameron Buie Dec 7 '12 at 4:28

The confusion maybe because every subgroup of a free abelian group is free abelian, while for the quotient group this is not necessarily true. The canoical example maybe $\mathbb{Z}/p\mathbb{Z}$, where $p$ is prime. Here $\mathbb{Z},p\mathbb{Z}$ are both free but the above group is not free. The wiki article probably can provide more information on this.

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The fact that you mention is a more general fact, i.e. every module is factor of a free module - you can imagine the construction in the following way:

take "enough" generators and create a free module over them, then, since different modules differ in "which elements are the same", i.e. in relations between the elements (therefore between the generators), create a submodule of relations $R$; factoring $R$ away is just like saying "the relations in $R$ are not important, zero" and in this way you can naturally get any module.

The example of $\mathbb{Z}_p$'s is:

take one element $x$ and create free group, so you get $F = \mathbb{Z}x$ (which is isomorphic to $\mathbb{Z}$); now you want to say that if two elements differ in a multiple of $p$, they're the same, so the relation submodule $R$ is $p\mathbb{Z}x$ (it contains $...,-1p, 0x, px, 2px..$) and factoring it out you say that all these elements are equal to each other (so to $0$ as well). If now you have for example $(2p + r)x$ and $(3p + r)x$ two elements of $F$, their difference is $p$, which lies in $R$, i.e. it "is not important", these two are the same in factor and you got the structure of $\mathbb{Z}_p$.

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