Talk:Residually finite group

Unsourced (nontrivial?) theorem

An equivalent definition is that a group is residually finite if the intersection of all its finite index subgroups is trivial.

Er, really? Without thinking about this, at all, it is quite obvious that a group is RF if the intersection of all its finite index normal subgroups is trivial. Can someone provide a proof for the other equivalence?

RandomP 14:45, 22 September 2006 (UTC)[reply]

Probably an error? I changed it to "normal", which generally occurs in the definition of "residually something". - Momotaro 14:00, 21 February 2007 (UTC)[reply]

It wasn't an error: in any group, the intersection of all subgroups of finite index is the same as the intersection of all normal subgroups of finite index. (By the way, the merge tags you added have been there for more than a month, and nobody has disagreed, so I suggest we go ahead and merge. The merged article should be at residually finite group.) --Zundark 07:57, 3 April 2007 (UTC)[reply]

This is true, because any finite index subgroup contains a finite index normal subgroup: A finite index subgroup has only finitely many subgroups conjugate to itself. Hence the intersection of all subgroups conjugate to a given finite index subgroup is of finite index; it is obviously normal, too. Thus the intersection of of finite index subgroups is contains the intersection of all f.i. normal subgroups. The other inclusion is obvious, I guess.138.246.7.146 (talk) 11:24, 11 November 2009 (UTC)[reply]