# Abstract Algebra Questions .

• jbarrera
In summary, the first problem shows that the union of two subgroups of a group is closed under inverses. The second problem proves that the centralizer of an element in a group is also a subgroup of that group.

#### jbarrera

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.
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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.

what you did on 2 is fine. you could have saved a little time by showing b-1 is in Z(g) whenever b is, and then showing ab-1 is in Z(g) when a and b are, but not much.

for 1) x in HUK means:

x is in H...or
x is in K..or both.

so start by assuming x is in H, what can you say about x-1?

next, if x is not in H, it must be in K, and use a similar agument.