Second Isomorphism Theorem: Statement, Proof

In this article, we will learn about the second isomorphism theorem for groups. This is also known as the diamond isomorphism theorem. The statement with its proof is provided below.

Second Isomorphism Theorem Statement

Let G be a group such that

1. H be a subgroup of G
2. K be a normal subgroup of G.

Then we have a group isomorphism

H/(H∩K) ≅ HK/K.

Second Isomorphism Theorem Proof

The below steps have to be followed to prove the second isomorphism theorem for groups.

Step 1: To show HK is a subgroup of G.

In order to prove this, we need to show that HK=KH. For h ∈ H, k ∈ K, as K is normal in G, we have that hkh-1 ∈ K.

∴ hk = (hkh-1)h ∈ KH.

Hence, HK ⊆ KH.

In a similar way, KH ⊆ HK.

Therefore HK = KH, proving that HK is a subgroup of G.

Step 2: To show K is normal in HK.

As HK is a subgroup of G by Step 1, and K is normal in G by assumption, one can easily deduce that K is a normal subgroup of HK.

Step 3: To show H∩K $\trianglelefteq$ H. This follows from the fact that K $\trianglelefteq$ G.

We will now prove the isomorphism.

Step 4: Define a mapping

φ: H → HK/K

by φ(h) = hK for h ∈ H.

Note that φ(h1h2) = h1h2K = (h1K) (h2K) = φ(h1) φ(h2), so φ is a group homomorphism.

By the definition of φ, φ is onto.

Now, Ker φ = {h ∈ H: φ(h) = K}

= {h ∈ H: hK = K}

= {h ∈ H: h ∈ K}

= H∩K.

So by the first isomorphism theorem of groups, we can conclude that

H/(H∩K) ≅ HK/K.

This proves the second isomorphism theorem for groups.

FAQs on Second Isomorphism Theorem

Q1: What is second isomorphism theorem for groups?

Answer: Let G be a group and H, K be its two subgroups. If K is normal in G, then we have a group isomorphism H/(H∩K) ≅ HK/K.

Q2: What is first isomorphism theorem for groups?

Answer: Let G and H be two groups and Let φ: G → H be an onto homomorphism. Then we have a group isomorphism G/ker(φ) ≅ H.

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