# Third Isomorphism Theorem: Statement, Proof

On this page, we will learn about the third isomorphism theorem for groups along with its statement and proof.

## Third Isomorphism Theorem Statement

Let G be a group and H, K be its two normal subgroups such that H ≤ K. Then we have a group isomorphism

(G/H)/(K/H) ≅ G/K.

## Third Isomorphism Theorem Proof

First, we prove that K/H is a normal subgroup of G/H. For gH∈ G/H and kH ∈ K/H, we have that

gH kH (gH)-1

= gH kH g-1 H

= gkg-1 H

∈ K/H as gkg-1∈K. This is because K is normal in G.

This shows that K/H is normal in G/H.

Next, we prove the group isomorphism. Let us define a mapping

φ: G/H → G/K

by φ(gH) = gK for g ∈ G.

Let us show that φ is well-defined. For g1, g2∈G with g1H=g2H, we have that

g1 = g2h for some element h∈H.

As H ≤ K, we have h∈K. Thus, we can conclude that

g1K=g2K

⇒ φ(g1H) = φ(g2H).

This shows that φ is well-defined.

By definition, φ is onto.

Now, Ker φ = {gH∈ G/H: φ(gH) = K}

= {gH∈ G/H: gK = K}

= {gH∈ G/H: g ∈ K}

= G/K.

Hence, by the first isomorphism theorem of groups, we obtain that

$\dfrac{G/H}{K/H} \cong G/K$.

This completes the proof of the third isomorphism theorem for groups.

For Other Isomorphism Theorems Click Below:

First Isomorphism Theorem

Second Isomorphism Theorem