Abstract algebra subgroup proofs..

[trickslapper]

1.Let n>1 be an integer, and let a be a fixed integer. Prove or disprove that the set

H={x?Z|ax ==0 (mod n)}

is a subgroup of Z under addition.

*== stands for "is congruent to"

2.Let H be a subgroup of G, let a be a fixed element of G, and let K be the set of all elements of the form aha[sup:14zijsu5]-1[/sup:14zijsu5], where h ? H. That is,

K={x ? G|x=aha[sup:14zijsu5]-1[/sup:14zijsu5] for some h ? H}.

Prove or disprove that K is a subgroup of G.


3. Prove or disprove that H ={h?G|h[sup:14zijsu5]-1[/sup:14zijsu5]=h} is a subgroup of the group G if G is abelian.


I know that i have to show the three parts of being a subgroup: closed, has an identity element, and has inverses. But i'm not sure how to do this with these problems, if someone could show me how to do even one of these i would greatly appreciate it.

thanks!

 

[daon]

I'll do one, you try to follow my lead:

1) 0 belongs to H since a*0=0. Hence H is nonempty.
2) Suppose h, g belongs to H. Then a*h = nk and a*g=nl for some k,l (why?). Therefore a*hg = (a*h)*g = n(kg), so that hg belongs to H, and hence H is closed.
3) if h belongs to H then a*h = nk and thus a*(-h) = n(-k) so -h belongs to H.

Therefore, H is a subgroup of Z.

 

[trickslapper]

Ok i understand you're example im gonna try #2 now..

1. Show k is not empty.

Since H is a subgroup of G it must have an identity element e and this implies that aea[sup:afmy2az5]-1[/sup:afmy2az5] is in K, so K is not empty.

2. Show that k is closed (x in K and y in K implies that xy in K)

Let x,y exist in K. Then x=ah[sub:afmy2az5]1[/sub:afmy2az5]a[sup:afmy2az5]-1[/sup:afmy2az5] for some h[sub:afmy2az5]1[/sub:afmy2az5] in H
and y=ah[sub:afmy2az5]2[/sub:afmy2az5]a[sup:afmy2az5]-1[/sup:afmy2az5] for some h[sub:afmy2az5]2[/sub:afmy2az5]in H

Then xy=(ah[sub:afmy2az5]1[/sub:afmy2az5]a[sup:afmy2az5]-1[/sup:afmy2az5])(ah[sub:afmy2az5]2[/sub:afmy2az5]a[sup:afmy2az5]-1[/sup:afmy2az5])= ah[sub:afmy2az5]1[/sub:afmy2az5](a[sup:afmy2az5]-1[/sup:afmy2az5]a)h[sub:afmy2az5]2[/sub:afmy2az5]a[sup:afmy2az5]-1[/sup:afmy2az5]

=a(h[sub:afmy2az5]1[/sub:afmy2az5]h[sub:afmy2az5]2[/sub:afmy2az5])a[sup:afmy2az5]-1[/sup:afmy2az5], and h[sub:afmy2az5]1[/sub:afmy2az5]h[sub:afmy2az5]2[/sub:afmy2az5] are in H

this implies that xy is in K and so K is closed under whatever operation

3. K contains inverses (x in K implies x[sup:afmy2az5]-1[/sup:afmy2az5] in K)

Let x exist in K, again this implies x=aha[sup:afmy2az5]-1[/sup:afmy2az5] for some h in H

take the inverse of both sides and you get: x[sup:afmy2az5]-1[/sup:afmy2az5]= (aha[sup:afmy2az5]-1[/sup:afmy2az5])[sup:afmy2az5]-1[/sup:afmy2az5], and by the reverse order law this is the same as:

(a[sup:afmy2az5]-1[/sup:afmy2az5])[sup:afmy2az5]-1[/sup:afmy2az5] h[sup:afmy2az5]-1[/sup:afmy2az5] a[sup:afmy2az5]-1[/sup:afmy2az5] = ah[sup:afmy2az5]-1[/sup:afmy2az5]a[sup:afmy2az5]-1[/sup:afmy2az5], where h[sup:afmy2az5]-1[/sup:afmy2az5] is in H

and this implies that x[sup:afmy2az5]-1[/sup:afmy2az5] is In K

so all three conditions are met and so K is a subgroup of G


I went to my professor and he kinda led me to this path is it correct?

 

[daon]

Good! It looks like you are getting the hang of it. But if you allow me to nitpick:

1. Show k is not empty.

Since H is a subgroup of G it must have an identity element e and this implies that aea[sup:em4kffbg]-1[/sup:em4kffbg] is in K, so K is not empty.

This isn't completely wrong, but it might be better stated as: e is in H since it is a subgroup, and e=eee[sup:em4kffbg]-1[/sup:em4kffbg], hence e belongs to K. The issue is, you don't know there is anything besides e in H to begin with, so presuming there is some "a" floating around makes this awkward.

(a[sup:em4kffbg]-1[/sup:em4kffbg])[sup:em4kffbg]-1[/sup:em4kffbg] h[sup:em4kffbg]-1[/sup:em4kffbg] a[sup:em4kffbg]-1[/sup:em4kffbg] = ah[sup:em4kffbg]-1[/sup:em4kffbg]a[sup:em4kffbg]-1[/sup:em4kffbg], where h[sup:em4kffbg]-1[/sup:em4kffbg] is in H

and this implies that x[sup:em4kffbg]-1[/sup:em4kffbg] is In K

By the way, this operation (really an action) is called conjugation, and it shows up A LOT.

edited my edit brain fart :lol:

 

[trickslapper]

Ok yea , that makes sense, I'm starting to like abstract algebra a lot more... you wouldn't happen to be awesome at real analysis would you haha?

 

[trickslapper]

I got another one i'm working and i think i have it 2/3 of the way done but i can't get the last part (show that x[sup:9toenep8]-1[/sup:9toenep8] exists)

For fixed integers a and b, let S={ax+by|x?Z and y?Z}.

Prove that S is a sugbroup of Z under addition.

1. 0?Z and 1?Z so for fixed intergers a and b, a(0)+b(1) gets into S so S is not empty.

2. Let p and q ? S, then p = ax[sub:9toenep8]1[/sub:9toenep8]+by[sub:9toenep8]1[/sub:9toenep8] and q=ax[sub:9toenep8]2[/sub:9toenep8]+by[sub:9toenep8]2[/sub:9toenep8]

after that all i did was add them (since addition is our operation) and show that pq ? S

3. I'm stuck here, i don't know for sure how to show that if p ? S then p[sup:9toenep8]-1[/sup:9toenep8]?S

any ideas? (also i'm not so sure on the first two parts, but they make sense to me)


Edit: would it be ok to say that p[sup:9toenep8]-1[/sup:9toenep8]=(ax+by)[sup:9toenep8]-1[/sup:9toenep8]=a(-x)+b(-y), where -x,-y ? Z and this implies that p[sup:9toenep8]-1[/sup:9toenep8] ? S?

 

[daon]

I got another one i'm working and i think i have it 2/3 of the way done but i can't get the last part (show that x[sup:yqmoo87c]-1[/sup:yqmoo87c] exists)

For fixed integers a and b, let S={ax+by|x?Z and y?Z}.

Prove that S is a sugbroup of Z under addition.

1. 0?Z and 1?Z so for fixed intergers a and b, a(0)+b(1) gets into S so S is not empty.

2. Let p and q ? S, then p = ax[sub:yqmoo87c]1[/sub:yqmoo87c]+by[sub:yqmoo87c]1[/sub:yqmoo87c] and q=ax[sub:yqmoo87c]2[/sub:yqmoo87c]+by[sub:yqmoo87c]2[/sub:yqmoo87c]

after that all i did was add them (since addition is our operation) and show that pq ? S

3. I'm stuck here, i don't know for sure how to show that if p ? S then p[sup:yqmoo87c]-1[/sup:yqmoo87c]?S

any ideas? (also i'm not so sure on the first two parts, but they make sense to me)


Edit: would it be ok to say that p[sup:yqmoo87c]-1[/sup:yqmoo87c]=(ax+by)[sup:yqmoo87c]-1[/sup:yqmoo87c]=a(-x)+b(-y), where -x,-y ? Z and this implies that p[sup:yqmoo87c]-1[/sup:yqmoo87c] ? S?

The notation is a little weird here, since you are mixing the notation from multiplicative groups with additive groups. I have seen it "defined" this way, so perhaps its just my personal prefernce. You essntially have it: If ax+by belongs to S, then since x,y are integers so are -1*x=-x, -1*y=-y. Hence a(-x)+b(-y) belongs to S and a(-x)+b(-y) = -(ax+by), which is the inverse of (ax+by).