
#1
Nov111, 08:58 PM

P: 3

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. 



#2
Nov211, 02:48 PM

Sci Advisor
P: 906

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. 


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