Question #17575

Let G^n be the smallest subgroup of a group G that contains the set {g^n, g element of G).
Describe elements of G^n.
Prove G^n is a normal subgroup of G.
Prove that for all x element of G/G^n, x^n = e.

Expert's answer

1) The subgroup G^n consists of finite products

g_1^n * g_2^n *... * g_k^n

for some elements g_1,g_2,...g_k from G.

2) Suppose q belongs G^n, so

q = g_1^n * g_2^n * ... * g_k^n

and x belogns to G.

We should prove that x^{-1} q x belongs to G^n.

First notice that for any x and g

x^{-1} g^n x =x^{-1} g g ... g x =

=x^{-1} g x x^{-1} g x ... x^{-1} g x =

=(x^{-1} g x)^n.

Hence

x^{-1} q x = x^{-1} g_1^n g_2^n ... g_k^n x =

= x^{-1} g_1^n x x^{-1} g_2^n x ... x^{-1} g_k^n x =

= (x^{-1} g_1 x)^n (x^{-1} g_2 x)^n ... (x^{-1} g_kx)^n and so x^{-1} q x belongs to G^n as well.

3) Let xG^n be an adjacency class of the factor groupG/G^n for some x from G.

Then

( xG^n )^n =(x^n) G^n,

but by definition x^n belongs to G^n, whence (x^n) G^n = G^n is a trivial class, that is ( xG^n )^n = (x^n) G^n = G^n = e the unit ofG/G^n.

g_1^n * g_2^n *... * g_k^n

for some elements g_1,g_2,...g_k from G.

2) Suppose q belongs G^n, so

q = g_1^n * g_2^n * ... * g_k^n

and x belogns to G.

We should prove that x^{-1} q x belongs to G^n.

First notice that for any x and g

x^{-1} g^n x =x^{-1} g g ... g x =

=x^{-1} g x x^{-1} g x ... x^{-1} g x =

=(x^{-1} g x)^n.

Hence

x^{-1} q x = x^{-1} g_1^n g_2^n ... g_k^n x =

= x^{-1} g_1^n x x^{-1} g_2^n x ... x^{-1} g_k^n x =

= (x^{-1} g_1 x)^n (x^{-1} g_2 x)^n ... (x^{-1} g_kx)^n and so x^{-1} q x belongs to G^n as well.

3) Let xG^n be an adjacency class of the factor groupG/G^n for some x from G.

Then

( xG^n )^n =(x^n) G^n,

but by definition x^n belongs to G^n, whence (x^n) G^n = G^n is a trivial class, that is ( xG^n )^n = (x^n) G^n = G^n = e the unit ofG/G^n.

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