# A subgroup intersects every conjugacy class

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

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