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Classify abelian groups $A$ which are irreducible $End(A)$-modules.

I think i did it for finite abelian group $A$ .

A finite abelian group $A$ is irreducible iff order of $A$ a is power of prime. What about infinite case i think prüfer groups , $\mathbb{Z}$ and $\mathbb{Q}$ is not irreducible but i can't give any general statement for infinite abelian group.Please give hint .

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    $\begingroup$ Your classification for finite abelian groups is wrong. $F_p^2$ is an irreducible$ GL_2 (F_p)$ module $\endgroup$
    – hunter
    Commented May 2, 2014 at 17:27
  • $\begingroup$ @hunter Yes you are right i think it should be power of prime . $\endgroup$
    – bytrz
    Commented May 2, 2014 at 17:34
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    $\begingroup$ Even so, your classification is incomplete, for if $A$ is a (finite or not) abelian group and $n\geq 2$ is an integer, then $A_n=\lbrace a\in A\mid na=0\rbrace$ is an $\mathrm{End}(A)$-submodule, in particular, if $A$ has order a primer power, and the maximal order of an element is $>p$ (so that it is $p^N$ for some integer $N\geq 2$), then for all $0<n<N$, $A_{p^n}$ is a nontrivial $\mathrm{End}(A)$ submodule of $A$. For example $\Bbb Z/p^N\Bbb Z$ for $N>1$ has prime order yet isn't irreducible. $\endgroup$ Commented May 2, 2014 at 17:44
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    $\begingroup$ ^ In conclusion, the only finite abelian groups which are simple over their endomorphism ring are elementary abelian. $\endgroup$
    – anon
    Commented May 2, 2014 at 17:46
  • $\begingroup$ @OlivierBégassat and sea turtles thanks for your comment they are important for me to see my mistakes. $\endgroup$
    – bytrz
    Commented May 2, 2014 at 18:07

1 Answer 1

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Hint. For any integer $n\geq 2$, there are $\mathrm{End}(A)$-stable subgroups $$A_n=\lbrace\,a\in A\mid na=0\,\rbrace$$ and $$nA=\lbrace\,a\in A\mid \exists\alpha\in A,\,a=n\alpha\,\rbrace\,.$$ Edit. I don't have a complete answer for when $A$ is infinite, but, assuming the axiom of choice, the answer is: $A$ is irreducible iff it has a vector space structure (i.e. $A$ is the underlying group of some vector space). There are two cases for any $\mathrm{End}(A)$-irreducible abelian group $A\neq 0$.

  1. Either, for all prime numbers $p$, $A_p=0$, $pA=A$, and $A$ is a $\Bbb Q$-vector space, and $\mathrm{End}_{Ab}(A)=\mathrm{End}_{\Bbb Q}(A)$. If $A$ is finite dimensional, then $A$ will be $\mathrm{End}_{Ab}(A)$ irreducible without further assumptions, indeed, if we fix $a\in A\setminus\lbrace0\rbrace$ and any $a'\in A$, there exists many $\Bbb Q$-linear maps sending $a$ to $a'$. The same holds true if $A$ has a $\Bbb Q$-basis, for instance if we assume the axiom of choice.
  2. Otherwise, there exists a single prime number $p$ with $A_p=A$ and $pA=0$, with for all other primes $q$, $A_q=0$ and $qA=A$. $A$ is then naturally a $\Bbb Z/p\Bbb Z$-vector space, and, again, $\mathrm{End}_{Ab}(A)=\mathrm{End}_{\Bbb Z/p\Bbb Z}(A)$. If $A$ is finite, then $A\simeq(\Bbb Z/p\Bbb Z)^n$ is irreducible by the same vector space argument as above, while if $A$ is infinite, the answer will likely depend on some form of the axiom of choice. If $A$ is known to have a basis, it certainly is true that it will be irreducible over $\mathrm{End}_{Ab}(A)$.
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  • $\begingroup$ @Oliver i think first one is for finite abelian groups. And for infinite case if i want A be irreducible then $A=nA$ so for example $Q=nQ$ so can i say Q is irreducible ? Is it enought to say A is irreducible iff $nA=A$ ? $\endgroup$
    – bytrz
    Commented May 2, 2014 at 18:36
  • $\begingroup$ $\Bbb Q$ is clearly irreducible, since there are endomorphisms "$r\times-$" of $\Bbb Q$ as an abelian group (i.e. as a $\Bbb Q$-vector space, see above) given by multiplication by any rational number $r\in \Bbb Q$. So the smallest $\mathrm{End}_{Ab}(\Bbb Q)$-stable subgroup of $\Bbb Q$ containing $r_0\neq 0$ is equal to all of $\Bbb A$, since we can always send $r_0$ to an arbitrary $r\in\Bbb Q$ by the map "$\frac{r}{r_0}\times-$". $\endgroup$ Commented May 2, 2014 at 18:50
  • $\begingroup$ However the Prüfer group $G=G(p)$ (for some prime $p$) isn't irreducible, as the subgroup of $p$-th roots of unity is stable under all endomorphisms of $G$, since it is equal to $p_G$, yet it is neither $0$ nor $G$. $\endgroup$ Commented May 2, 2014 at 18:53
  • $\begingroup$ Also the integers $\Bbb Z$ aren't irreducible, since $n\Bbb Z$ for $n\geq 2$ is a stable subgroup that isn't $0$ nor the whole of $\Bbb Z$. $\endgroup$ Commented May 2, 2014 at 18:54
  • $\begingroup$ @OlivierBégassat i can't understand 1. at your hint . How you say $A$ is $Q$ vector space and what is the mean of \mathrm{End}_{Ab} ? $\endgroup$
    – bytrz
    Commented May 2, 2014 at 19:18

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