Abstract AlgebraHomomorphisms and Isomorphisms of GroupsIsomorphism and Correspondence Theorems
Second Isomorphism Theorem
After using normal subgroups to describe homomorphic images, we now compare a subgroup with a quotient involving a normal subgroup. The focus keyword second isomorphism theorem refers to the isomorphism H/(H∩K)≅HK/K, where H≤G and K⊴G. The theorem says that when H is combined with K, the part of H already lying inside K is exactly the part that disappears in the quotient. Students often find the notation dense, but the idea is simple: first restrict to H, then collapse the overlap with K.
:::definition[Product of Subgroups]
Let H and K be subgroups of a group G. The product set HK is defined by
HK={hk∣h∈H,k∈K}.
If K⊴G, then HK is a subgroup of G.
:::
:::theorem[Second Isomorphism Theorem]
Let H≤G and let K⊴G. Then H∩K⊴H, K⊴HK, and
H/(H∩K)≅HK/K.
:::
:::proof
Given that H≤G and K⊴G.
To prove that H∩K⊴H, K⊴HK, and H/(H∩K)≅HK/K.
Since K⊴G, the product HK is a subgroup of G, and K⊴HK.
First prove H∩K⊴H. Let x∈H∩K and h∈H. Since x,h∈H, we have hxh−1∈H. Since x∈K and K⊴G, we also have hxh−1∈K. Therefore hxh−1∈H∩K. Hence H∩K⊴H.
Define ϕ:H→HK/K by
ϕ(h)=hK.
[1] To prove that ϕ is a homomorphism.
For h1,h2∈H,
ϕ(h1h2)ϕ(h1)ϕ(h2)=h1h2K,=(h1K)(h2K)=h1h2K.
Thus ϕ(h1h2)=ϕ(h1)ϕ(h2).
[2] To compute the kernel.
kerϕ={h∈H∣hK=K}={h∈H∣h∈K}=H∩K.
[3] To compute the image.
Every element of HK/K has the form hkK with h∈H and k∈K. Since kK=K, we have hkK=hK=ϕ(h). Therefore ϕ(H)=HK/K.
By the first isomorphism theorem,
H/kerϕ≅ϕ(H).
Substituting kerϕ=H∩K and ϕ(H)=HK/K gives
H/(H∩K)≅HK/K.□
:::
The quotient HK/K records the part added to K by elements of H. However, any element of H already lying in K becomes invisible after quotienting by K. That invisible part is precisely H∩K. Therefore H/(H∩K) and HK/K carry the same information. The second isomorphism theorem is often called the diamond theorem because the subgroups H, K, H∩K, and HK form a useful diagram of inclusions.
:::example[Integer Subgroups]
Let H=(2) and K=(3) in (Z,+). Since Z is abelian, every subgroup is normal. Here
H+K=(2)+(3)=Z
and
H∩K=(6).
By the second isomorphism theorem,
(2)/(6)≅Z/(3).
Thus (2)/(6) has the same group structure as Z3.
:::
This preview specializes the second isomorphism theorem to subgroups (m) and (n) of Z. The top of the diamond is the sum (m)+(n)=(gcd(m,n)), and the bottom is the intersection (m)∩(n)=(lcm(m,n)). Adjust m and n and observe how the overlap changes. The two side quotients have the same number of cosets, which reflects the isomorphism. Try m=4,n=6 and then m=2,n=3 to compare a nontrivial overlap with the example above.
:::scientific-preview[Integer Diamond Explorer]
Second Isomorphism Theorem | BMLabs | Homomorphisms and Isomorphisms of Groups | BMLabs Mathematics | BMLabs Mathematics
Visual Learning UG
Abstract Algebra · Homomorphisms and Isomorphisms of Groups
Second Isomorphism Theorem
Learn Second Isomorphism Theorem. This page develops the main mathematical ideas in a clear sequence.
After using normal subgroups to describe homomorphic images, we now compare a subgroup with a quotient involving a normal subgroup. The focus keyword second isomorphism theorem refers to the isomorphism H/(H∩K)≅HK/K, where H≤G and K⊴G. The theorem says that when H is combined with K, the part of H already lying inside K is exactly the part that disappears in the quotient. Students often find the notation dense, but the idea is simple: first restrict to H, then collapse the overlap with K.
Core definition02
Product of Subgroups
Let H and K be subgroups of a group G. The product set HK is defined by
HK={hk∣h∈H,k∈K}.
If K⊴G, then HK is a subgroup of G.
Key result03
Second Isomorphism Theorem
Let H≤G and let K⊴G. Then H∩K⊴H, K⊴HK, and
H/(H∩K)≅HK/K.
Reasoning pathway04
Given that H≤G and K⊴G.
To prove that H∩K⊴H, K⊴HK, and H/(H∩K)≅HK/K.
Since K⊴G, the product HK is a subgroup of G, and K⊴HK.
First prove H∩K⊴H. Let x∈H∩K and h∈H. Since x,h∈H, we have hxh−1∈H. Since x∈K and K⊴G, we also have hxh−1∈K. Therefore hxh−1∈H∩K. Hence H∩K⊴H.
Define ϕ:H→HK/K by
ϕ(h)=hK.
[1] To prove that ϕ is a homomorphism.
For h1,h2∈H,
Thus ϕ(h1h2)=ϕ(h1)ϕ(h2).
[2] To compute the kernel.
kerϕ={h∈H∣hK=K}={h∈H∣h∈K}=H∩K.
[3] To compute the image.
Every element of HK/K has the form hkK with h∈H and k∈K. Since kK=K, we have hkK=hK=ϕ(h). Therefore ϕ(H)=HK/K.
By the first isomorphism theorem,
H/kerϕ≅ϕ(H).
Substituting kerϕ=H∩K and ϕ(H)=HK/K gives
H/(H∩K)≅HK/K.
□
The quotient HK/K records the part added to K by elements of H. However, any element of H already lying in K becomes invisible after quotienting by K. That invisible part is precisely H∩K. Therefore H/(H∩K) and HK/K carry the same information. The second isomorphism theorem is often called the diamond theorem because the subgroups H, K, H∩K, and HK form a useful diagram of inclusions.
Guided example06
Integer Subgroups
Let H=(2) and K=(3) in (Z,+). Since Z is abelian, every subgroup is normal. Here
H+K=(2)+(3)=Z
and
H∩K=(6).
By the second isomorphism theorem,
(2)/(6)≅Z/(3).
Thus (2)/(6) has the same group structure as Z3.
This preview specializes the second isomorphism theorem to subgroups (m) and (n) of Z. The top of the diamond is the sum (m)+(n)=(gcd(m,n)), and the bottom is the intersection (m)∩(n)=(lcm(m,n)). Adjust m and n and observe how the overlap changes. The two side quotients have the same number of cosets, which reflects the isomorphism. Try m=4,n=6 and then m=2,n=3 to compare a nontrivial overlap with the example above.
Visual laboratory
Integer Diamond Explorer
INTEGER DIAMOND EXPLORER
Dynamic Sandbox
Initializing Workspace
The lower node is what H and K share, and the upper node is what they generate together. The theorem compares moving from H down to the overlap with moving from H+K down to K. In the integer case, the matching quotient orders make the abstract isomorphism visible.
This calculator verifies the second isomorphism theorem for integer subgroups. Enter positive integers m and n. It computes (m)+(n) using the greatest common divisor and (m)∩(n) using the least common multiple. The two quotient orders should match every time, because (m)/(lcm(m,n)) and (gcd(m,n))/(n) describe the same quotient structure.
Interactive calculator
GCD LCM Isomorphism Calculator
GCD LCM ISOMORPHISM CALCULATOR
Initializing Workspace
The numerator on the left is H, so only the portion of K inside H can disappear there. On the right, quotienting H+K by all of K removes the same information. That is why the intersection appears in one denominator and K appears in the other.
Worked problem12
Let H=(m) and K=(n) be subgroups of (Z,+). Describe H+K and H∩K.
Complete solution13
Let H=(m) and K=(n) be subgroups of Z.
The subgroup generated by both m and n is generated by their greatest common divisor. Hence
(m)+(n)=(gcd(m,n)).
The intersection consists of all integers divisible by both m and n. Hence it is generated by the least common multiple:
(m)∩(n)=(lcm(m,n)).
Therefore the second isomorphism theorem gives
(m)/(lcm(m,n))≅(gcd(m,n))/(n),
where the quotient on the right is interpreted inside (gcd(m,n)).
□
Worked problem14
Let H≤G and K⊴G. Prove directly that H∩K⊴H.
Complete solution15
Let x∈H∩K and let h∈H.
Since x∈H and h∈H, the subgroup property of H gives
hxh−1∈H.
Since x∈K and K⊴G, conjugation by any element of G, and therefore by h, keeps x inside K. Hence
hxh−1∈K.
Thus hxh−1∈H∩K.
Hence H∩K⊴H.
□
Learning tip16
In the second isomorphism theorem, the kernel of the map h↦hK is not all of K; it is the part of K that lies inside H. That is why the denominator is H∩K.
Independent practice17
In Z, apply the second isomorphism theorem to H=(4) and K=(6).
If H∩K={e}, what does the theorem say?
If H⊆K, what does the theorem reduce to?
Answer18
H+K=(2) and H∩K=(12), so (4)/(12)≅(2)/(6).
It gives H≅HK/K.
If H⊆K, then HK=K and H∩K=H, so both quotient groups are trivial.
Questions to consolidate
Frequently Asked Questions
3
1Why must K be normal?
Normality ensures HK is a subgroup and that HK/K is a quotient group.
2Is H required to be normal in G?
No. The theorem only needs H≤G and K⊴G.
3What is the main map in the proof?
It is ϕ:H→HK/K defined by ϕ(h)=hK.
Continue learning
Continue to the Third Isomorphism Theorem
Learn how quotienting in two stages gives the same structure as quotienting directly by a larger normal subgroup.
