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If the subgroup is normal, I understand why it is true, but if the subgroup isn't normal I'm not sure why a conjugacy class couldn't get larger once you get the $ghg^{-1}$ elements

Eric Stchiike
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    In most cases, the conjugacy class for the whole group is not even contained in the subgroup; it would be more reasonable to ask whether a conjugacy class of a subgroup is the intersection with the subgroup of a conjugacy class of the whole group. But this is not true in general either, and not even (contrary to what you say in the question) for normal subgroups. – Marc van Leeuwen Apr 02 '15 at 11:20

2 Answers2

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A conjugacy class of a subgroup need not of course be a conjugacy class of the whole group. For instance, every nonabelian group has abelian subgroups (ie cyclic subgroups). In those abelian subgroups, the conjugacy classes are all singleton sets (they only have one element), but in the original group, only the elements of the center form conjugacy classes of only one element...

Theo Douvropoulos
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    Thanks! I'm reading my notes on representation theory and we introduced a way to induce characters, we have a class function $f$ on H and we define $\dot{f}$ a class function on G this way: $\dot{f}(g)=f(g)$ if $g\in H$ and 0 otherwise. It will not be a class function of G then! I knew something was wrong. – Eric Stchiike Apr 02 '15 at 03:56
  • Indeed, you'll need $H$ to be normal. Glad to help – Theo Douvropoulos Apr 02 '15 at 03:59
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Hint: What happens if the subgroup is abelian?

Mariano Suárez-Álvarez
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