17
$\begingroup$

Can monoids of endomorphisms of nonisomorphic groups be isomorphic ?

$\endgroup$
  • $\begingroup$ The monoid of continuous endomorphisms of the circle group is isomorphic to the integers under multiplication, which is of course also the endomorphism monoid of $\mathbb Z$. (There are loads of discontinuous ones though, so that doesn't help much.) $\endgroup$ – lambda Nov 12 at 13:08
  • 7
    $\begingroup$ I think one has $\mathrm{Aut}(\mathrm{SL}_2(p))=\mathrm{Aut}(\mathrm{PSL}_2(p))=\mathrm{PGL}_2(p)$ for every prime $p\ge 5$, and the endomorphism monoid is just the same adding the constant $1$. For $p=5$ this gives the alternating group $\mathrm{Alt}_5$ vs the binary icosahedral group, which have isomorphic endomorphism monoids. $\endgroup$ – YCor Nov 12 at 13:36
17
$\begingroup$

For any prime $p$, the endomorphism monoid of $\mathbb{Z}[\frac{1}{p}]$ is a commutative monoid with zero whose submonoid of nonzero elements is the direct sum of a cyclic group of order two (generated by multiplication by $-1$), an infinite cyclic group (generated by multiplication by $p$), and a free commutative monoid on countably many generators (multiplication by other primes).

But different primes give nonisomorphic groups.

$\endgroup$
  • 4
    $\begingroup$ Just an obvious remark: the endomorphism monoid of the group $A=\mathbf{Z}[1/p]$ (or more generally of any unital subring $A$ of $\mathbf{Q}$) is canonically isomorphic to the multiplicative monoid $(A,\times)$. Which indeed for $\mathbf{Z}[1/p]$ has the given description (and more generally for $\mathbf{Z}[S^{-1}]$ for a set of primes $S$, its isomorphism type as monoid retains only the cardinal of $S$ and of its complement). $\endgroup$ – YCor Nov 12 at 15:17
15
$\begingroup$

It is proved in $[$1$]$ that the tetrahedral group $A_4$ of order $12$ and the binary tetrahedral group of order $24$ have isomorphic endomorphism monoids. So this gives a finite example. It is also the smallest order example. Moreover, no other finite groups have isomorphic endomorphism monoid to the endomorphism monoid of these two groups without being isomorphic to one of them.

$[$1$]$ P. Puusemp, Groups of order 24 and their endomorphism semigroups, J. Math. Sci. (2007) 144(2) 3980–3992 (link at Springerlink).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.