Let G be a group and let H be a subgroup of G a Show that fo

Let G be a group and let H be a subgroup of G. (a) Show that for any g G, the set gHg¹ = {ghg¹ |h H} is also a subgroup of G. (b) Show that if |H| = m and H is the only subgroup of G with order m, then H is normal in G.

Solution

Given that G is a group and H a subgroup of G.

a) Let g be any element in G.

The subset gHg^-1 consists of all elements of the form {ghg1 |h H}

To prove that gHg^-1 is a subgroup.

Closure:

Consider two elements h1 and h2in H

gh1g¹*gh2g¹ = gh1(g¹*g)h2g¹

= gh1eh2g¹

=  g(h1h2)g¹

Since H is a subgroup h1h2 is in H say h\'

=  gh\'g¹

Hence closure is true.

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Identity element is geg^-1 =e belongs to gHg^-1

Inverse for ghg^-1 is gh^-1g^-1

So gHg^-1 is a subgroup of G

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b) If o(H) = m

no of elements in H are m.

Since there is only one subgroup with order m,

gHg^-1 is a subgroup of G with order m.

So gHg^-1 has to be same H as there is only one subgroup with order m.

Since gHg^-1=H,

H is normal in G