alext87 Posted May 11, 2008 Share Posted May 11, 2008 Is this correct? Let G be a group and H a subgroup of G. Suppose that G acts on the (non-empty) set A. Then there exists an element of A which has H as its stabilizer? The converse is easily shown. As if G acts on a set A and then pick any w belonging to A then the the stabilizer of w is a subgroup of G. I.e. it there a one-one correspondence between the subgroups of G and the stabilizers of the elements of A? Any help would be great cheer.s Link to comment Share on other sites More sharing options...
Al Don Gate Posted July 3, 2008 Share Posted July 3, 2008 Hello, I hope i have understood your question... Let be the multiplicative group G={-1,1} acting on itself by translation : [math](g,x)\mapsto gx[/math] If i look at the sub group G (which is well a sub group of G), G can't be written as th stabilizater of a element of G because : Stab(1)={1} Stab(-1)={1} But maybe you think at a stric subgroup of G... Hi, Another example more significant (i hope) : Let be [math]G=\mathbb{Z}/4\mathbb{Z}(=\{\overline{0},\overline{1},\overline{2},\overline{3}\})[/math] the additive subgroup of congruence modulo 4. We see that [math]H=\{\overline{0},\overline{2}\}[/math] is a subgroup of G. Suppose G act on itself by translation. Then there is no element of G which have H for stabilizer. And here H is a strict subgroup. Link to comment Share on other sites More sharing options...
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