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Assume $B,H,G$ are abelian groups, $f:B\rightarrow H$ is a surjective homomorphism, $H$ is a subgroup of $G$. My question is :is there an abelian group $A$ and a surjective homomorphism $g:A\rightarrow G$ satisfies:

(1) $B$ is a subgroup of $A$.

(2) $g\upharpoonright B=f$.

(3) $\ker(g)=\ker(f)$.

Thank you for your answer.

Li Cong-xiao
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  • It's better to saying: $B$ is isomorphic by a subgroup of $A$. – hamid kamali May 06 '15 at 10:26
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    Were there a distinguished, nice representative $x$ of every coset $xH\in G/H$, then you can set $A=B\times G/H$ with $g(b, xH)= f(b)x$. This works, for instance, for $B=\Bbb R^n, G=\Bbb R^m$ and $H=\Bbb R^r$ with $r$ small enough, and $f$ being orthogonal projection onto some $r$-dimensional subspace (the distinguished representatives are the points orthogonal to $H$ in $G$). Not sure how to get such a distinguished representative in the general case (when orthogonality makes no sense). – Arthur May 06 '15 at 10:57
  • Addendum: If $G\approx G'\times H$ for some $G'$, it works nicely. For other cases, try $G=\Bbb Z$ and $H=n\Bbb Z$, since that's the simplest case where it's not so. – Arthur May 06 '15 at 11:10
  • Arthur, the case that $H$ is a direct summand of $G$ is OK. But how to handle the general case? – Li Cong-xiao May 06 '15 at 11:33
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    @user73985 I don't think this works because from the condition (1),(2),(3) you must have $|A|=|B|\times |G/H|$ and here $|A|=|B|\times |G|/|B|=|G|$. – Clément Guérin May 06 '15 at 11:46
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    @ClémentGuérin ah yes, and the issue is that $B$ is not a subgroup of $A$ – Christopher May 06 '15 at 11:54

1 Answers1

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I think this works but it is a little complicated (too much but I cannot see a faster way).

  1. I claim that I only have to do it for abelian $p$-groups. So I take $p$ a prime numbers and assume that $G,H,B$ are abelian $p$-groups.

    1. I claim that there exists between $H$ and $G$ a sequence of $p$-groups :

$$H:=H_0\subset H_1\subset ... \subset H_r:=G $$

With $|H_i|/|H_{i-1}|=p$. Now if I can do the extension when $G/H$ is of cardinal $p$, I can extend from $H_0$ to $H_1$ then from $H_1$ to $H_2$... from $H_{r-1}$ to $G$. So it suffices to do it when $G/H$ is of cardinal $p$. This is my assumption now.

Take $x\in G$ such that $x\notin H$ then, because $G/H$ is cyclic of cardinal $p$, any element $g$ of $G$ can uniquely be written :

$$g=hx^l\text{ where } h\in H\text{ and } 0\leq l\leq p-1 $$

Furthermore (again because the quotient is cyclic of order $p$), $x^p\in H$. Take $y\in B$ such that $f(y)=x^p$. Now as a set we define $A:=B\times \mathbb{Z}_p$, any element of $A$ is written as :

$$(b,l)\text{ where } b\in B\text{ and } 0\leq l\leq p-1 $$

Define your function $g$ as :

$$g:B\times\mathbb{Z}_p\rightarrow G $$

$$(b,l)\mapsto f(b)x^l $$

This is a surjective function. I claim that we can choose a group law on $A$ that makes $A$ an abelian group and $g$ a group morphism. Indeed on $A$ set :

$$(b_1,l_1).(b_2,l_2):=(b_1b_2,l_1+l_2)\text{ if } l_1+l_2<p $$

$$(b_1,l_1).(b_2,l_2):=(b_1b_2y,l_1+l_2-p)\text{ if } l_1+l_2\geq p $$

It is obviously an internal law, commutative with a neutral element. The only thing to verify is that this is associative. Well I leave this to you (it is a carefull study but this works, actually one can sees that changing $B$ into $H$ you get the law of $G$). Now it is obvious (almost) to see that $g$ is then a group morphism which verifies all the properties.

Clément Guérin
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  • How to construct $A$ and $g$ when $B$, $H$, $G$ are all torsion-free? – Li Cong-xiao May 06 '15 at 12:21
  • @LiCong-xiao, I am sorry, I had in my mind finite abelian groups... For the infinite group case I have no idea, I will think about this but I think this is not true, if I find anaything I will edit my question. – Clément Guérin May 06 '15 at 12:27
  • Thank you for your answer for the finite groups. – Li Cong-xiao May 06 '15 at 12:33
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    @LiCong-xiao, the question you asked is solved in Fuchs'book "Infinite abelian groups". The proposition 24.6 says that this is always true. From what I have read the argument is to include the group $G$ in a divisible group. I will let you read it for further details (this is not trivial). My answer works (with a little work for the generalization) anytime the quotient $G/H$ is finitely generated (essentially because it works for cyclic quotients, hence for finite groups, and it works when the quotient is isomorphic to $\mathbb{Z}$ because $\mathbb{Z}$ is projective)... – Clément Guérin May 12 '15 at 07:56
  • Oh, it's right, I have checked the detail following Fuchs' book. Your answer is a more operable method-- it is difficult to understand why put the G in a divisible group in Fuchs' book. – Li Cong-xiao May 15 '15 at 04:21