Proof: G/H1 is Isomorphic to H2/K for G with Normal Subgroups H1 and H2

[Poirot1]

Let
G be a group with normal subgroups H1 and H2 with H2 not a subset of H1. Let K = H1 intersect H2.


Show that if G/H1 is simple, then G/H1 is isomorphic to H2/K.


My first thought was to set up a homomorphism with K as the kernel but soon realized that the fact that H2 was not normal is H1 scuppered this tactic. G/H1 being simple implies that H1 is the largest proper normal subgroup but where to go from there?

 

[Poirot1]

Just realized that my last sentence is incorrect. G/H1 being simple means there is no normal subgroup A of G which H1 is normal in.

 

[Opalg]

The quotient map $\pi:G\to G/H_1$ maps $H_2$ to a normal subgroup of $G/H_1$. This normal subgroup contains more than just the identity element, so by simplicity it must be the whole of $G/H_1$. Now show that the kernel of the homomorphism $\pi|_{H_2}$ is equal to $K$.

 

[Poirot1]

The quotient map $\pi:G\to G/H_1$ maps $H_2$ to a normal subgroup of $G/H_1$. This normal subgroup contains more than just the identity element, so by simplicity it must be the whole of $G/H_1$. Now show that the kernel of the homomorphism $\pi|_{H_2}$ is equal to $K$.

Ah, my first ideas were correct.

 

[blue_raver22]

Thank you for your question. It is always exciting to see someone exploring the intricacies of group theory.

First, let's define some terms for clarity:

- A group G is said to be simple if it has no proper nontrivial normal subgroups, meaning that the only normal subgroups of G are the trivial subgroup {e} and the whole group G itself.

- A normal subgroup H of a group G is a subgroup that is invariant under conjugation, meaning that for any element g in G, the conjugate of H by g (denoted by gHg^-1) is also a subgroup of G.

- The notation G/H denotes the quotient group of G by the normal subgroup H, where the elements of G/H are the cosets of H in G and the group operation is defined as (aH)(bH) = (ab)H for all a, b in G.

Now, let's consider the given situation: G is a group with normal subgroups H1 and H2, where H2 is not a subset of H1. We also have K = H1 ∩ H2, the intersection of H1 and H2.

We want to show that if G/H1 is simple, then G/H1 is isomorphic to H2/K.

To do this, we will use the First Isomorphism Theorem, which states that if φ: G → G' is a group homomorphism with kernel K, then G/K is isomorphic to φ(G) (the image of G under φ).

In our case, we can define a homomorphism φ: G → H2/K by sending each element g in G to its coset gK in H2/K. This is indeed a homomorphism because for any two elements g1 and g2 in G, we have φ(g1g2) = (g1g2)K = (g1K)(g2K) = φ(g1)φ(g2).

Now, let's look at the kernel of φ. If g is in the kernel of φ, then φ(g) = gK = K, which means that g is in K. But K = H1 ∩ H2, so g is in both H1 and H2. This means that g is in the intersection of H1 and H2, which is K. Therefore, the kernel of φ is K.

Since H2 is