Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Subgroups?
Prove or disprove that H (the set of elements h that are a member of G, such that h^-1 = h) is a subgroup if the group G is abelian.
So clearly there exists an inverse for each element h in H since h^-1 = h.
But how do we show whether H has an identity and is closed?
h.h^-1 = e = h.h^-1 = h.h
Of course, h.h = e so h.h is a member of H and this is closure. But how do I show that h.h exists in H?
3 Answers
- az_lenderLv 76 years ago
The identity is in H because the identity HAS the property of being its own inverse!
To show that this subset of G is closed is a little more complicated. Suppose h1*h1 = 1 and h2*h2 = 1, but h1 is not h2. Then how do we know the product h1*h2 is in H? Well, because its inverse is h2^(-1)*h1^(-1) = h2*h1, and if G is abelian, then h2*h1 = h1*h2, so h1*h2 is the inverse of h1*h2, so h1*h2 is in H. That proves the closure, and also shows why the restriction of G being abelian was necessary.
- ?Lv 76 years ago
Consider the subset of G = G / K where K = kernel. What can you surmise about h in G / K? What can you surmise of h in K?
- 6 years ago
(Identity) ee = e ⇒ e∈H
(Inverses) ∀ h∈H, hh = e ⇒ h^-1∈H
(Associativity) ∀ a,b,c∈H, a(bc) = (ab)c (since a,b,c∈H⊆G)
(Closure) ∀ a,b∈H, (ab)(ab) = (ab)(ba) (G is abelian) = a(bb)a = aea = aa = e ⇒ ab = (ab)^-1 ⇒ ab∈H
