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Normal subgroup proof

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Let G be a finite group, let N be a normal subgroup of G, and let x be an
element of G. Show that if the order of x in G is relatively prime to
|G|/|N|, then x is an element of N.

We know that xNx^(-1) is identical to N when N is normal, for any x.
Also we know that |G|/|N| is a factor of (or divides) |G|.
How to show x in G is also in N?

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Solution Summary

This is a proof regarding a normal subgroup.

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Proof:
Since N is normal in G, we have a projection f: G->G/N such that f(g)=gN, for any g in G.
Now x is an element in G, f(x)=xN, then f(<x>)=<xN>, where <x> is the ...

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