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Abstract Algebra Questions...
I have two problems that I'm a little puzzled by, hopefully someone can shed some light.
1) Show that if H and K are subgroups of the group G, then H U K is closed under inverses.
2) Let G be a group, and let g ε G. Define the centralizer, Z(g) of g in G to be the subset
Z(g) = {x ε G | xg = gx}.
Prove that Z(g) is a subgroup of G.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For problem 2 this is what I have but I am not sure if it is correct.
Since eg = ge for g in G, we know Z(g) is not the empty set.
-Take a in Z(g) and b in Z(g), and take any g in G, then we have...
(ab)g = a(bg) = a(gb) = (ag)b = (ga)b = g(ab). Thus ab is in Z(g).
- Take a in Z(g) and g in G. Then we know...
ag = ga
(a^-1* a )g = (a^-1 * g) a (multiplying both sides by a inverse)
e * g = a^-1 * g*a
g * a^-1 = a^-1 * g * (a * a^-1) ( multiplying again by a invese)
g * a^-1 = a^-1 * g
Thus a^-1 is in Z(g), so Z(g) is a subgroup of G.
I have two problems that I'm a little puzzled by, hopefully someone can shed some light.
1) Show that if H and K are subgroups of the group G, then H U K is closed under inverses.
2) Let G be a group, and let g ε G. Define the centralizer, Z(g) of g in G to be the subset
Z(g) = {x ε G | xg = gx}.
Prove that Z(g) is a subgroup of G.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For problem 2 this is what I have but I am not sure if it is correct.
Since eg = ge for g in G, we know Z(g) is not the empty set.
-Take a in Z(g) and b in Z(g), and take any g in G, then we have...
(ab)g = a(bg) = a(gb) = (ag)b = (ga)b = g(ab). Thus ab is in Z(g).
- Take a in Z(g) and g in G. Then we know...
ag = ga
(a^-1* a )g = (a^-1 * g) a (multiplying both sides by a inverse)
e * g = a^-1 * g*a
g * a^-1 = a^-1 * g * (a * a^-1) ( multiplying again by a invese)
g * a^-1 = a^-1 * g
Thus a^-1 is in Z(g), so Z(g) is a subgroup of G.