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?
az_lender
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.
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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?
Mike
(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