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In the book Character Theory Of Finite Groups by I.Martin Issacs as exercise 2.14

Let $G$ be a finite group with commutator subgroup $G'$. Let $H \subset G' \cap Z(G)$ be cyclic of order n and let m be the maximum of the orders of elements of G/H. Assume that n is a prime power and show that $|G| \geq n²m$.

I have tried, following the hints given, to show, taking $\chi$ to be an irreducible character whose kernel intersects trivially with $H$, that $\chi(1) \geq n$. But, thus far as I am concerned, I see no idea of how to show the inequality, not to mention the latter part of the proof, as suggested, to deploy the Problem 2.9(b) which states that $\chi(1) \leq |G:A|$ for any abelian subgroup A of G.
I have asked my teacher, who replied that one is to show that $|G/H| \geq nm$, of which the proof I have no idea either. Of course, if one can demonstrate the existence of one subgroup of $G/H$ order $\geq nm$, then the problem should be resolved, as the teacher suggested. But this still bewilders me at present.

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P.S. This question is found at the page 31. –  awllower May 30 '12 at 7:57
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It was not clear to me from what you wrote whether you had succeeded in proving that $\chi(1) \geq n$ for a character of the type you considered, or whether this is a homework problem. This is the case. Consider an element of $H$ of order $n.$ Let $\sigma$ be an irreducible complex representation affording a character $\chi,$ and assume that $H \cap {\rm ker} \sigma = 1.$ THen $h\sigma$ is a matrix of order $n,$ but must also be a scalar matrix since $\sigma$ is irreducible and $H \leq Z(G).$ Furthermore, ${\rm det} \sigma$ is a $1$-dimensional representation of $G,$ so contains $G^{\prime}$ in its kernel. Since $H \leq G^{\prime},$ we can now conclude that $h\sigma$ is a scalar matrix of order $n$ and determinant $1.$ Let $h \sigma = \omega I$ for some primitive $n$-th root of unity $\omega.$ Then ${\rm det}(h\sigma) = \omega^{\chi(1)}.$ Hence $\omega^{\chi(1)} = 1$ and $\chi(1)$ must be divisible by $n.$ In particular, $\chi(1) \geq n.$ Now let $g \in G$ be an element such that $gH$ has order $m$ in G/H. Let $L = \langle g,H \rangle.$ Then $H \leq Z(L)$ and $L/H$ is cyclic, so $L$ is Abelian, and has order $mn.$ Hence with our character $\chi$ as before, we now have $n \leq \chi(1) \leq [G:L].$ Thus $|G| \geq n|L| \geq n^{2}m.$ I don't think the assumption that $n$ is a prime power is necessary, in fact.

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Thanks a lot!! Indeed the part that $\omega$ is of order n, and hence n divides $\chi(1)$ is excellent!! As the book describes, the assumption on n can be removed after further developpments, while I think it is not necessary now. –  awllower May 30 '12 at 9:01
    
The assumption on n, however, turns out to be necessary. Indeed, as $H$ is then a $p$-group, and it is cyclic, we can conclude that there is a faithful irreducible character $\chi$ of $H$, and then by reciprocity find an irreducible character of $G$, whose restriction to $H$ becomes$\chi$, thus obtaining an irreducible character of $G$ which intersects trivially with $H$. Per chance there is still some other way to arrive at the same conclusion? Thanks again. –  awllower May 31 '12 at 8:11
    
I do not understand what you are saying in the two comments above. The inequality in the problem as you originally asked is indeed true whether or not $n$ is a prime power. In any case, any finite cyclic group has a faithful irreducible necessarily degree $1$), whether or not its order is a prime power. If $H = \langle h \rangle$, where $h$ has order $n,$ then define $\sigma : H \to \mathbb{C}$ via $h^{j}\sigma = e^{\frac{2 \pi i j}{n}}.$ –  Geoff Robinson May 31 '12 at 9:05
    
Thanks again. I see the reason for that characters now. –  awllower Jun 1 '12 at 7:03
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