Let H and K be two subgroups of a group G. We define a HK as a subset of G as HK
ID: 3404124 • Letter: L
Question
Let H and K be two subgroups of a group G. We define a HK as a subset of G as HK = {hk|h H, k K}. Show that if HK = G, H K = {e} and hk = kh, h H, k K, then H × K is isomorphic to G. Let H and K be two subgroups of a group G. We define a HK as a subset of G as HK = {hk|h H, k K}. Show that if HK = G, H K = {e} and hk = kh, h H, k K, then H × K is isomorphic to G. Let H and K be two subgroups of a group G. We define a HK as a subset of G as HK = {hk|h H, k K}. Show that if HK = G, H K = {e} and hk = kh, h H, k K, then H × K is isomorphic to G.Explanation / Answer
First show that hkh-1k-1=e for all hH, kK. Then show that the function
f:H×KG given by f(h, k) =hk is a group isomorphism.
For all hH, kK, we have
hkh-1k-1= (hkh-1)k-1=h(kh-1k-1)HK
so hkh-1k-1=e
. This means that hk=kh
Now prove that f is an isomorphism,
For h, h'H and k, k'K, we compute:
f(h, k)·f(h', k') = (hk)(h'k')
= h(kh')k'
= h(h'k)k'
= (hh')(kk')
= f(hh', kk')
= f((h, k)·(h', k'))
Hence f is a homomorphism. Since HK=G, any element gG
may be written as g=hk for some hH, kK, and
f(h, k) =hk = g, so f is surjective. Finally, if (h, k)ker f, then hk=e, so
h=k-1HK={e}, so (h, k) = (e, e). Hence f is injective.
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