Definition: Let G be a group with operation *. we say H is a subgroup of G if
(i) Ø ≠ H ⊂ G,
(ii) g * h ∈ H for all g, h ∈ H, and
(iii) gi ∈ H for all g ∈ H. (gi is the inverse of g).
(a) Prove that if H is a subgroup of G then e ∈ H. (e is the identity.)
(b) Suppose that for each a ∈ A, Ha is a subgroup of G. Prove that K = ∩ Ha is a subgroup of G.
a ∈ A
(c) The trivial group is {e}. Find examples of non-trivial, proper subgroups of (ℤ,+) such that the intersection of two of them is an non-trivial subgroup. Find examples of two subgroups whose union is not a subgroup.
I've been stuck on this forever. I can't figure it out. I sorta proved the first part. Please help. Thanks so much!!
(i) Ø ≠ H ⊂ G,
(ii) g * h ∈ H for all g, h ∈ H, and
(iii) gi ∈ H for all g ∈ H. (gi is the inverse of g).
(a) Prove that if H is a subgroup of G then e ∈ H. (e is the identity.)
(b) Suppose that for each a ∈ A, Ha is a subgroup of G. Prove that K = ∩ Ha is a subgroup of G.
a ∈ A
(c) The trivial group is {e}. Find examples of non-trivial, proper subgroups of (ℤ,+) such that the intersection of two of them is an non-trivial subgroup. Find examples of two subgroups whose union is not a subgroup.
I've been stuck on this forever. I can't figure it out. I sorta proved the first part. Please help. Thanks so much!!
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(a) by (i) we have that H is non-empty, so there is some g in H.
by (iii) we have that g^-1 is also in H.
by (ii) we have that g*g^-1 is in H.
but (in G, and therefore also in H), g*g^-1 = e, so e is in H.
(b) let k, k' be any two elements of K (if K has only one element,
then K = {e}, since e is in every Ha). we need to show 2 things:
1- that k*k' is in K
2- that k^-1 is in K.
but k and k' are in every Ha, so in any one of those, k*k' is in that same Ha
since Ha is a subgroup. hence k*k' is in every Ha, so k*k' is in K.
a similar argument shows k^-1 must be in every Ha, so k^-1 is in K.
(c) let H = (2Z,+) = {2k : k in Z}, and let K = (3k,+) = {3k: k in Z}.
the intersection H∩K is 6Z, which is a non-trivial subgroup.
we can use these same 2 subgroups for the second part:
HUK is not a subgroup, because 2 is in HUK, and 3 is in HUK,
but 2+3 = 5 is not (5 is neither a multiple of 2 nor a multiple of 3).
by (iii) we have that g^-1 is also in H.
by (ii) we have that g*g^-1 is in H.
but (in G, and therefore also in H), g*g^-1 = e, so e is in H.
(b) let k, k' be any two elements of K (if K has only one element,
then K = {e}, since e is in every Ha). we need to show 2 things:
1- that k*k' is in K
2- that k^-1 is in K.
but k and k' are in every Ha, so in any one of those, k*k' is in that same Ha
since Ha is a subgroup. hence k*k' is in every Ha, so k*k' is in K.
a similar argument shows k^-1 must be in every Ha, so k^-1 is in K.
(c) let H = (2Z,+) = {2k : k in Z}, and let K = (3k,+) = {3k: k in Z}.
the intersection H∩K is 6Z, which is a non-trivial subgroup.
we can use these same 2 subgroups for the second part:
HUK is not a subgroup, because 2 is in HUK, and 3 is in HUK,
but 2+3 = 5 is not (5 is neither a multiple of 2 nor a multiple of 3).