proof of second isomorphism theorem for groups

First, we shall prove that HK is a subgroupMathworldPlanetmathPlanetmath of G: Since eH and eK, clearly e=e2HK. Take h1,h2H,k1,k2K. Clearly h1k1,h2k2HK. Further,


Since K is a normal subgroupMathworldPlanetmath of G and h2G, then h2-1k1h2K. Therefore h1h2(h2-1k1h2)k2HK, so HK is closed under multiplicationPlanetmathPlanetmath.

Also, (hk)-1HK for hH, kK, since


and hk-1h-1K since K is a normal subgroup of G. So HK is closed under inversesMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, and is thus a subgroup of G.

Since HK is a subgroup of G, the normality of K in HK follows immediately from the normality of K in G.

Clearly HK is a subgroup of G, since it is the intersectionMathworldPlanetmathPlanetmath of two subgroups of G.

Finally, define ϕ:HHK/K by ϕ(h)=hK. We claim that ϕ is a surjective homomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath from H to HK/K. Let h0k0K be some element of HK/K; since k0K, then h0k0K=h0K, and ϕ(h0)=h0K. Now


and if hK=K, then we must have hK. So


Thus, since ϕ(H)=HK/K and kerϕ=HK, by the First Isomorphism TheoremPlanetmathPlanetmath we see that HK is normal in H and that there is a canonical isomorphism between H/(HK) and HK/K.

Title proof of second isomorphism theorem for groups
Canonical name ProofOfSecondIsomorphismTheoremForGroups
Date of creation 2013-03-22 12:49:47
Last modified on 2013-03-22 12:49:47
Owner yark (2760)
Last modified by yark (2760)
Numerical id 17
Author yark (2760)
Entry type Proof
Classification msc 20A05
Related topic ProofOfSecondIsomorphismTheoremForRings