alext87 Posted May 11, 2008 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
Al Don Gate Posted July 3, 2008 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.
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