about cosets in abstract algebra?

Question

(1) let H be a subset of G. show that the following are equivalent.
i. a element bH
ii. b element aH
iii. aH=bH
iv. b^-1a element H
v. a^-1 element H

1. i iff ii
2. i iff iii
3. iv iff v
4. i iff v

(2) let G be a group of order pq, p and q are prime. show that every subgroup of G is cyclic.

Answer 1

You have a few typos:

H should be a subgroup, of course - that's how you form cosets. I assume this is what you meant, and won't berate you about it like some answerers.

"v. a^-1 b element H"

is what the fifth statement should say, I think.

#1

i → ii

Take a ∈ bH. Then a = bh for some h ∈ H. Take h' ∈ H to be the inverse of h. Then we know ah' = bhh' = be = b. Thus b ∈ aH.

ii → i

Proof is exactly the same as i→ii, since a and b are generic elements of the group G.

i → iii

Take a ∈ bH. But a∈ aH since a = ae and e∈ H. Thus bH ⊂ aH.

On the other hand, a and b are generic elements, so we can also infer by similar argument (or by application of i→ii) that aH ⊂ bH.

Thus aH = bH.

iii → i

Proof is trivial. If aH = bH, then a ∈ aH = bH and vice versa automatically.

iv → v

If b' a ∈ H, then b'a = h for some h∈H. This means that a = ea = bb'a = bh for some h ∈ H. Thus a ∈ bH.

v → iv

Take a∈ bH. Then a = bh for some h ∈ H. This means that b'a = b'bh = h for some h ∈ H. Thus b'a ∈ H.

i → v

Take a'b ∈ H. Then a'b = h for some h ∈ H. Then we know b = eb = aa'b = ah, which means b ∈ aH. Since ii → i, we conclude a ∈ bH.

v → i

Take a'b ∈ H. Then a'b = h for some h ∈ H. That means b = eb = aa'b = ah. Thus b ∈ aH, and since ii→ i, we know a ∈ bH too.

#2

Here you must mean proper subgroups, or else this statement is not true (since G itself is not necessarily cyclic). This is either a typo, or is implied by the question (since the improper subgroup is invariably uninteresting).

But take any proper subgroup H ⊂ G. We know |H| divides |G|, due to Lagrange's theorem, but |H|<|G|. There are only three possibilities:

|H| = p
|H| = q
|H| = 1

So either |H| is prime, or it's 1, and either way, we know this is a cyclic group. Any group of prime order is a cyclic group (and the group of order 1 is trivially cyclic).

Answer 2

I can show a case in which the above statements are not true, if H does not have to be a group.
Take the integers mod 17 under +. Then, take H = {3,4,5,6}, a = 6, b = 3. Then 6 is in {6, 7, 8, 9}, but 3 is not in {9, 10, 11, 12}. Obviously, this does not hold for subsets.

Assuming that H has to be a subgroup, which you did not state:

Now, I don't think that v is equivalent. Is it possible that is a typo? I will assume statement v is that a^-1 b is an element of H.
We can show the rest by showing that i => iv => ii => v => i, and these all imply iii, which implies one of these.
If a is in b H, then b^-1a is in H, by multiplication by b^-1. By this we also know that its inverse, a^-1b is in H, so b is in a H. By the same argument, ii => v => i.

Now, since a^-1b is an element of H, a^-1 b H is contained in or equal to H, and therefore, a a^-1 b H is contained in or equal to a H, but that is just the same as b H contained in or equal to a H. By symmetry, a H is contained in or = b H

If a H = b H, then b^-1 a H = H, but e is an element of H if it is a group. Therefore, a^-1 b e is equal to an element of H, so a^-1 b is an element of H.

Answer 3

Some problems here:
(1). Do you want H to be a subset of G or a subgroup of G?
(2). Here you must say "every proper subgroup".
The whole group may not be cyclic, for example
Z_3 x Z_3 is a noncyclic group of order 9.
No. 2 is easy: The order of any proper subgroup H must
divide pq by Lagrange's theorem.
So the order of H must be p or q. Thus H is a
group of prime order, so it is cyclic.

Let's do part of no. 1, assuming H is a subgroup of G.
i iff ii.
Let a ε bH
Then a = bh for h ε H.
Thus b = ah^-1.
Since h is a subgroup of G, h^-1 ε H
So b ε aH.
I don't see how to do this if h^-1 is not in H.
To go the other way,
let b = ha and note that
a = bh^-1,
so a ε bH.
I'll let you work out the rest. Again assume
H is a subgroup of G in all cases.
Good luck!

Answer 4

i implies ii:

if a is in bH, then a = bh, where h is some element of H. But in any group every element, in particular h, has an inverse, so call it h^-1. Multiply on the right by this and we get ah^-1 = b, hence b equals a times some element in H (since H is a subgroup, it contains h^-1). So then b is in aH, and we are done.

to show ii implies i, the same argument works with a and b interchanged, so infact we have proven both. The rest of the proofs are very similar and you will learn more by trying them. Good luck, and email me if you get stuck - i can tell you faster by chatting.

The p and q problem is true by Lagrange's Theorem. It says that the order (size) of any subgroup divides the oreder of the whole group. See if you can figure it out now. Again if you get stuck I can help you more. Good luck.

Answer 5

John's answer is very exhaustive, mine is shorter: sure this is an "iff" (if and provided that) fact. for coset multiplication to artwork, we choose each and every left coset of H to be a acceptable coset (of the comparable representatives) as nicely. John's answer has the tedious information. (surely, if we wish (aH)(bH) = abH, then this suggests ahbh' = abh", so hbh' = bh" so hb = bh"h'^-a million. if H isn't prevalent, then there'll be some b for which Hb ? bH, so there'll be some h with hb not in bH, yet for coset multiplication to artwork, hb has to equivalent bh"h'^-a million that's clearly a area of bH). in spite of the incontrovertible fact that, you're in success: {?^0 = e,?^2} is unquestionably a common subgroup; enable's look into the cosets: H = {e,?^2} ?H = {?,?^3} ?H = {?,??^2} ??H = {??,??^3} the suitable cosets are the comparable: H = {e.?^2} H? = {?,?^3} = ?H H? = {?,?^2?} = {?,??^2} = ?H by way of fact; ?^2? = ?(??) = ?(??^-a million) = ?(??^3) = (??)?^3 = (??^-a million)?^3 = ??^2 H?? = {??,?^2??} = {??,??^3} = ??H by way of fact: ?^2(??) = (??^2)? = (??^2)? (see above) = ??^3. so H is prevalent. now we are able to easily sq. each and every coset: (H)(H) = H (by way of fact H is a subgroup, it extremely is often real). (?H)(?H) = ?^2H = H (by way of fact ?^2H = H, provided that ?^2 is in H). (?H)(?H) = ?^2H = H (provided that ?^2 = e). (??H)(??H) = (??)^2H = H (by way of fact (??)^2 = e: (??)^2 = (??)(??) = ?(??)? = ?(??^-a million)? = ?^2 = e). provided that H is the id of G/H.....