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I believe that it is true that if we have a group $G$, and two copies $H_1$, $H_2$ of some group $H$ as subgroups of $G$, we can fix a representation $V$ of $H$ and have the situation:

$$\operatorname{Ind}_{H_{1}} V \ncong \operatorname{Ind}_{H_{2}} V$$

I believe my example is Klein-4 in $S_4$, you can take a normal copy generated by products of disjoint $2$-cycles $(12)(34)$, etc., and a copy $\{ (12)(34),(12),(34), 1 \}$ (or do $\mathbb{Z}_3 \times \mathbb{Z}_3$ in $S_9$ )

Anyway, if $H_1$ and $H_2$ are conjugate subgroups, then the induced representations will be isomorphic as can be seen by the character formula

$ 1/|H| \sum_{x} V(x^{-1}gx) $ = character of $\operatorname{Ind}_{H_{1}} V $ = character of $\operatorname{Ind}_{H_{2}} V$

Is this correct? Thank you

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So you're looking at the case of ordinary representations? – Curufin Nov 30 '12 at 14:15
up vote 1 down vote accepted

Yes, if $H_{1}$ and $H_{2}$ are conjugate, you will get isomorphic modules (whihever ring you work over). If $H_{1}$ and $H_{2}$ are not conjugate, then you need not get the same character for the induced module, even in the complex case. The example you gave for $S_{4}$ is perfectly valid- if $H_{1}$ is normal, then the induced character will vanish outside $H_{1},$ while if $H_{2}$ is not normal, that need not be the case (it is not in your example).

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Thank you for the answer, it is what I thought. I wanted to ask here for a "fact-checking", in case I did something wrong in my proof. The reason why I picked those two versions of klein-4 is that one is normal and the other isn't ( when you unpack the character formula, all the terms in the normal case will remain, while the terms in the non-normal case will be 0, when conjugation sends you outside the subgroup ). – Elliot Nov 30 '12 at 15:34

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