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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

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Hello,

I hope i have understood your question...

Let be the multiplicative group G={-1,1} acting on itself by translation :

$(g,x)\mapsto gx$

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 $G=\mathbb{Z}/4\mathbb{Z}(=\{\overline{0},\overline{1},\overline{2},\overline{3}\})$ the additive subgroup of congruence modulo 4.

We see that $H=\{\overline{0},\overline{2}\}$ 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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