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For a subgroup $H$ of a given group $G$, I say $H$ is "big" if it has nonempty intersection with each conjugacy class of $G$. I have known that, trivially, $G$ itself is "big". And if $H$ is a normal subgroup and it is "big", then $H=G$. I have also known that a finite group has no proper "big" subgroup. My question is "Is there an infinite group who has a proper 'big' subgroup?"

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standard examples: max torus in cpt lie gp. or subgp of upper triangular matrices in gln(k), k alg closed. – Peter McNamara Apr 4 '12 at 3:33
The example of upper triangular matrix is so standard, thank you for your example! – Song Li Apr 4 '12 at 4:38
Just to make Peter's first example yet more concrete: all unitary matrices are diagonalizable. – Allen Knutson Apr 4 '12 at 7:07
This, by the way, may be the central point as to why representation theory of compact connected Lie groups is easier than that of finite groups; by character theory, a finite-dim complex representation of a group is determined up to isomorphism by its restriction to a "big" subgroup, and if that subgroup is abelian, so much the better! – Allen Knutson Apr 4 '12 at 7:09
Other standard example: $G$ is the group of bijections $\mathbb{Z} \to \mathbb{Z}$ which fix all but finitely many integers; $H$ is the subgroup of bijections fixing $0$. – David Speyer Apr 4 '12 at 14:01
up vote 12 down vote accepted

Yes, for example in Osin's infinite group with 2 conjugacy classes every proper subgroup is big. Of course if you do not care about the number of generators, you can consider the (much easier) infinitely generated group constructed by Higman-Neumann-Neumann where all non-identity elements are conjugate. There also every proper subgroup is big.

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Thank you for your answer. – Song Li Apr 4 '12 at 4:37
Every proper non-trivial subgroup is big. – Geoff Robinson Jul 21 '12 at 11:11
"Proper" includes "non-trivial". – Mark Sapir Jul 21 '12 at 12:06

How about for free groups?

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Yes, a free group seems to have a large subgroup. It may be constructed inductively adding $g$ to $<g_1,g_2,...,g_k>$ for each $g$ with $g^F\cap <g_1,...,g_k>=\emptyset$. (We need to start with a good initial $<g_1,g_2>$ to avoid getting all $F$.) – Lev Glebsky Jan 9 '13 at 0:29

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