Talk:Metacyclic group

Untitled
I think it might refer to the following paper

HANS J0RGEN MUNKHOLM, INDUCED MONOMIAL REPRESENTATIONS, YOUNG ELEMENTS, AND METACYCLIC GROUPS Proc. AMS 19 (1968), 453-458,

in which the author claimed that: "In [1, §28] Curtis and Reiner, after having constructed a full set of pairwise nonequivalent, irreducible left ideals of the group-algebras QSn by means of the so-called Young diagrams, raise the question whether the same set of ideas can be applied to other classes of finite groups to give every (isomorphism class of) irreducible representation. In this paper we give an affirmative answer to the question in the sense that we show that the idea can give the representations of the metacyclic groups."

The reference [1] is C. W. Curtis and Irving Reiner, Representation theory of finite groups and associative algebras, Pure and Appl. Math., vol 11, Interscience, New York, 1962. —Preceding unsigned comment added by 159.226.244.10 (talk) 02:44, 26 September 2010 (UTC)

Representations
Does anybody know about this 'easy construction of irreducible representations by diagrams'? It is certainly not in the only reference for this article. —Preceding unsigned comment added by 171.64.38.45 (talk) 04:48, 20 March 2009 (UTC)


 * It was by Zaremskym.  I think the irreducible representations should be fairly easy to find, especially for groups whose derived subgroup and whose maximal abelian quotient are both cyclic.  A couple of references where this is worked out explicitly are:
 * I don't have Grove's book here, but the second citation is fairly easy to read. It doesn't mention diagrams. JackSchmidt (talk) 14:45, 20 March 2009 (UTC)
 * Section 5 of has a description of the modular representations.  It is a little terse, but I think it just says they are very very clear.  He has a second paper,, that starts and ends with the metacyclic case, so might be easier to read.  I'll check it out when I check Grove.
 * (Never end with a list of data) JackSchmidt (talk) 15:11, 20 March 2009 (UTC)
 * Section 5 of has a description of the modular representations.  It is a little terse, but I think it just says they are very very clear.  He has a second paper,, that starts and ends with the metacyclic case, so might be easier to read.  I'll check it out when I check Grove.
 * (Never end with a list of data) JackSchmidt (talk) 15:11, 20 March 2009 (UTC)
 * (Never end with a list of data) JackSchmidt (talk) 15:11, 20 March 2009 (UTC)
 * (Never end with a list of data) JackSchmidt (talk) 15:11, 20 March 2009 (UTC)

Sorry, original commenter back. I agree that the irreducible representations are a fairly easy exercise for people who have taken courses in representation theory. The question only seems to be what 'diagrams' are useful in their computation. The papers referenced above don't seem to have diagrams, though of course they do get the representations. I'm deleting this for now, until someone else decides to revert or finds a source for that specific claim. —Preceding unsigned comment added by 128.253.227.65 (talk) 03:02, 12 July 2009 (UTC)

Finiteness Assumed?
Is there a tacit assumption that metabelian groups are finite? The notation $$C_m$$ and $$C_n$$ in the displayed sequence suggests this. On the other hand, if the definition is just an extension of a cyclic group by a cyclic group, then certainly Z could be used instead, no? Michael Kinyon (talk) 05:27, 14 July 2011 (UTC)


 * The notation used in the short exact sequence is misleading. There's no requirement for the cyclic groups to be finite. --Zundark (talk) 07:11, 14 July 2011 (UTC)


 * As I suspected. Thanks for fixing it.Michael Kinyon (talk) 21:54, 19 August 2011 (UTC)

A dicyclic group is not necessarily a semidirect product of two cyclic groups
The article (section Examples) says : "The direct product or semidirect product of two cyclic groups is metacyclic. These include the dihedral groups, the quasidihedral groups, and the dicyclic groups."

I think that a dicyclic group is not necessarily a semidirect product of two cyclic groups. Take the quaternion group. It is not cyclic, thus, if it was the semidirect product of two cyclic subgroups, these subgroups should be nontrivial. But the quaternion group is not the semidirect product of two nontrivial subgroups, because two nontrivial subgroups of the quaternion group have never a tivial intersection. (It is well known that the quaternion group has only one subgroup of order 2, thus this subgroup is contained in every nontrivial subgroup of the quaternion group.)

But, clearly, a dicyclic group has a cyclic subgroup of index 2, and is thus metacyclic. I suggest to write : "
 * The direct product or semidirect product of two cyclic groups is metacyclic. These include the dihedral groups and the quasidihedral groups.
 * The dicyclic groups are metacyclic. (Note that a dicyclic group is not necessarily a semidirect product of two cyclic groups.)" Marvoir (talk) 07:44, 3 December 2014 (UTC)