Hey flybynight.
What about associativity? Is there a condition with regards to (ab)^(-1) = b^(-1)*a^(-1)? Does this have implications on the associative axiom?
Hi Everyone,
My problem is stated as: "Determine the groups G such that is a subgroup on ." To me, appears that if G is a group, then this should always define a subgroup.
Going through the subgroup test:
Identity: All of the resulting subgroups will have an identity, namely .
Units: All elements are invertible. For any pair , the inverse is , which also exists in the subgroup.
Closed: For any , we would have , which should also exist in the subgroup.
Is there something I'm missing? The question seems to imply that in some cases, this subset will not be a subgroup, but I don't see when that would occur.
Thanks,
Peter
I promise I'm not being purposely dense. If I understand you correctly, we're dealing with something like: , which, when isolating the second element of each pair, can be grouped in two separate ways: or . But when doing the operation in G, there is no real difference between the operations, so far as I can tell.
What you're saying seems to be something like: consider and , which isolating the second term, gives and , respectively. However, the fact that these terms are different isn't problematic, so far as I can see. The answer will be different, but that makes sense, given that the starting elements are different. Unless, again, I'm confused.
I'm just giving some food for thought and not a solution: if my reasoning is wrong then like you, I would like to see where I have slipped up.
If the associativity fails then you don't have a group. So if (ab)^(-1)c^(-1) must equal a^(-1)(bc)^-1 then you don't have a group since a^(-1)(bc)^-1 = a^(-1)c^(-1)*b^(-1) and (ab)^(-1)c^(-1) = b^(-1)a^(-1)c^(-1) so you don't get an equality unless you impose very special conditions.
here is the problem:
it's NOT associativity:
[(g,g^{-1})*(h,h^{-1})]*(k,k^{-1}) =
(gh,g^{-1}h^{-1})*(k,k^{-1}) =
((gh)k,(g^{-1}h^{-1})k^{-1}) =
= (g(hk),g^{-1}(h^{-1}k^{-1})) (by associativity of the group product in G)
= (g,g^{-1})*[(h,h^{-1})*(k,k^{-1})]
it's NOT identity:
(g,g^{-1})*(e,e) = (ge,g^{-1}e) = (g,g^{-1}) (note that e^{-1} = e)
(e,e)*(g,g^{-1}) = (eg,eg^{-1}) = (g,g^{-1})
it's NOT inverses:
(g^{-1},g) is clearly a two-sided inverse for (g,g^{-1}).
hmm...well, then what is the problem?
the most BASIC fact of what a binary operation on a set MEANS: if a,b are in S, then a*b HAS TO BE IN S ALSO!
now (g,g^{-1})*(h,h^{-1}) = (gh,g^{-1}h^{-1})
ask yourself:
is g^{-1}h^{-1} equal to (gh)^{-1}? what do we call the groups in which this holds true?
What do you mean, vice versa? If the group is Abelian, then the equivalence holds? Also, I don't know if I am completely clear on why it is important that g^{-1}h^{-1}=(gh)^{-1}. If you have (g,g^{-1})*(h,h^{-1})=(gh,g^{-1}h^{-1}), g^{-1}h^{-1} should be in G, because G must include both inverses, and because G is closed, it must also include their product, right?
I really do appreciate you helping me,
Peter
yes G is closed, however the subset of GxG consisting of pairs (g,g^{-1}) may NOT be.
look, let's look at an example, so you can see why this fails:
let G = {e,a,a^{2},b,ab,a^{2}b} where a^{3} = b^{2} = e, and ba = a^{2}b.
the last rule: ba = a^{2}b let's us "move a's to the front", so which know WHICH element of G any product is. for example:
(ab)(a^{2}b) = a(ba)(ab) = a(a^{2}b)(ab) = (a^{3}b)(ab) = b(ab) = (ba)b = (a^{2}b)b = a^{2}b^{2} = a^{2}.
i'll list the inverses for you (you can verify them if you like):
e^{-1} = e (duh!)
a^{-1} = a^{2} (since a^{3} = e)
(a^{2})^{-1} = a
b^{-1} = b
(ab)^{-1} = ab (we can show this one of two ways: (ab)^{-1} = b^{-1}a^{-1} = ba^{2} = (ba)a = (a^{2}b)a = a^{2}(ba) = a^{2}(a^{2}b) = a^{4}b = ab, or:
(ab)(ab) = a(ba)b = a(a^{2}b)b = a^{3}b^{2} = ee = e, so ab is of order 2 and thus its own inverse)
(a^{2}b)^{-1} = a^{2}b
now the subset of GxG we are considering is the set:
H = {(e,e),(a,a^{2}),(a^{2},a),(b,b),(ab,ab),(a^{2}b,a^{2}b)}
let's look at:
(a,a^{2})*(ab,ab) = (a(ab),a^{2}(ab)) = (a^{2}b,a^{3}b) = (a^{2}b,b)
hmm...does it look like b is the inverse of a^{2}b? is the pair (a^{2}b,b) in H?
the closure of G only implies that the product of two elements (under * in GxG) is again in GxG. GxG has |G|^{2} elements. the subset:
H = {(g,h) in GxG: h = g^{-1}} only has |G| elements, it's a MUCH smaller set.