Characters of Normal Subgroup of Index 2

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Hi There,

Ok, I'm new to this so I'm sorry if this is abit warbled!...

We have a normal subgroup N of a finite group H such that [H:N]=2
We have a chatacter Chi belonging to the irreducible characters of H, Irr(H) , which is zero on H\N.

I have already shown that Chi restricted to N = Theta 1 + Theta 2, Where
Theta 1 is not equal to Theta 2, and both Theta 1, and Theta 2 belong to Irr(N).

I am now trying to show that given a conjugacy class K of H which is also a conjugacy class of N, we get Theta 1 (n) = Theta 2 (n) for n a member of K.

Intuitively, I can see this is true, I have seen examples in the alternating group A6, the normal subgroup of index 2 in S6 fro example.

I would be really greatful for a push in the right direction...I have books: 'James and Liebeck' and 'Isaacs'... if you could even direct me to some relevant info?

Wow, think my notation is abit crap, sorry.
 
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It is traditional (though not necessarily ideal) to use ascii-tex to write maths in electronic form if you're not going to use the LaTeX capabilities of this site.

Thus \theta_1 is written as theta_1 (or \theta_1).

_ means subscript and ^ means superscript.

Right. I don't know if you're supposed to do it this way but here's my idea.

Consider Ind_N^H(theta_1).
 
Thanks for the notation tip, don't know how to use LaTeX so will try the other method.
Ok, I'm looking at Ind_N^H(theta_1), using Frobenius Reciprocity, I have <Ind_N^H(theta_1) , Chi> = <theta_1, Chi_N> (where < , > denotes inner product)

Since Chi_N = theta_1 + theta_2 we get <Ind_N^H(theta_1) , Chi> = <theta_1, theta_1 + theta_2>

now, is <theta_1, theta_1 + theta_2> = <theta_1, theta_1> + < theta_1, theta_2> ? and where can i go from here to get to my result...as in have theta_1(n) = theta_2(n) for each n belonging to the conjugacy class K
 
It's an inner product - it is of course (conjugate) linear in each variable. You have possibly learned (and I was assuming you had) that irreducible characters form an orthonormal basis of the class functions. However, if you didn't know that <A+B,C>=<A,C>+<B,C> then I have perhaps made a mistake in where you've got to in learning the material.

To get the final result you just need to look at how to construct in induced character.
 
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The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
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