Help Peter Prove Simple Proposition on Simple Groups & Maximal Normal Subgroups

[Math Amateur]

I need help with the proof of an apparently simple Proposition in Aigli Papantonopoulou's book: Algebra: Pure and Applied.

The proposition in question is Proposition 5.2.3 and reads as follows:View attachment 3320Can someone please help me and provide an explicit and formal proof of this proposition.Since Proposition 5.1.3 is mentioned in the text above I am providing the statement and proof of this proposition as follows:View attachment 3321I would really appreciate some help with the above question.

Peter

 

[Euge]

I need help with the proof of an apparently simple Proposition in Aigli Papantonopoulou's book: Algebra: Pure and Applied.

The proposition in question is Proposition 5.2.3 and reads as follows:View attachment 3320Can someone please help me and provide an explicit and formal proof of this proposition.Since Proposition 5.1.3 is mentioned in the text above I am providing the statement and proof of this proposition as follows:View attachment 3321I would really appreciate some help with the above question.

Peter

Hi Peter,

Suppose $K$ is a maximal normal subgroup of $G$. Suppose $N$ is a nontrivial normal subgroup of $G/K$. Let $N'$ be the preimage of $N$ under the canonical projection $\pi : G \to G/K$. Then $N'$ is a normal subgroup of $G$ properly containing $K$. Since $K$ is maximal, it cannot be strictly contained in any proper subgroup of $G$. Therefore $N' = G$, and thus $N = G/K$. This shows that $G/K$ is simple.

Conversely, suppose $G/K$ is simple. Let $M$ be any subgroup of $G$ which properly contains $K$. Since $K$ is normal in $G$, it is also normal in $M$, and the quotient $M/K$ is normal in $G/K$. In addition, since $K \neq M$, $M/K$ is nontrivial. Therefore $M/K = G/K$ by simplicity of $G/K$. Hence $M = G$, and consequently $K$ is maximal.

 

[Math Amateur]

Hi Peter,

Suppose $K$ is a maximal normal subgroup of $G$. Suppose $N$ is a nontrivial normal subgroup of $G/K$. Let $N'$ be the preimage of $N$ under the canonical projection $\pi : G \to G/K$. Then $N'$ is a normal subgroup of $G$ properly containing $K$. Since $K$ is maximal, it cannot be strictly contained in any proper subgroup of $G$. Therefore $N' = G$, and thus $N = G/K$. This shows that $G/K$ is simple.

Conversely, suppose $G/K$ is simple. Let $M$ be any subgroup of $G$ which properly contains $K$. Since $K$ is normal in $G$, it is also normal in $M$, and the quotient $M/K$ is normal in $G/K$. In addition, since $K \neq M$, $M/K$ is nontrivial. Therefore $M/K = G/K$ by simplicity of $G/K$. Hence $M = G$, and consequently $K$ is maximal.

Thanks Euge ... working through your post now ...

Peter

 

[Deveno]

Proposition 5.1.3 is often called the lattice isomorphism theorem, because it establishes a lattice isomorphism between the subgroups of $G$ that contain $K$ and the subgroups of $\tau(G)$.

In these lattices, the partial order is set-inclusion, the JOIN of two subgroups $A,B$ is:

$\langle A,B\rangle$-the subgroup GENERATED by $A$ and $B$ (this is usually larger that the set-union of the subgroups involved).

and the MEET is: $A \cap B$.

The subgroups containing $K$ are also called "the subgroups over $K$".

It is enlightening to attempt to prove:

$\tau(\langle A,B\rangle) = \langle(\tau(A),\tau(B)\rangle$
$\tau(A \cap B) = \tau(A) \cap \tau(B)$.

Equivalently, one can show that:

$\tau(A) \subseteq \tau(B) \iff A \subseteq B$ (that is, $\tau$ induces an order-preserving map between these two lattices).

Often, it is only established in most group-theory texts that this is merely a bijection.

Rather than prove the lattice isomorphism theorem (it is straight-forward, and is a direct consequence of the Fundamental Isomorphism Theorem), I will illustrate it with an example:

Suppose $G = D_4$, the dihedral group of order 8. Recall that:

$D_4 = \langle r,s: r^4 = s^2 = 1, rs = sr^{-1}\rangle = \{1,r,r^2,r^3,s,sr,sr^2,sr^3\}$.

To specify a homomorphism $\phi: D_4 \to G'$, it suffices to specify $\phi(r)$, and $\phi(s)$, and to verify the relations of $D_4$ still hold.

We will now do so for $\phi:D_4 \to \Bbb Z_2 \times \Bbb Z_2$, by setting:

$\phi(r) = (1,0)$
$\phi(s) = (0,1)$.

Note that $4\phi(r) = 4(1,0) = (1,0) + (1,0) + (1,0) + (1,0) = (0,0) + (0,0) = (0,0)$, and:

$2\phi(s) = 2(0,1) = (0,1) + (0,1) = (0,0)$, while:

$\phi(rs) = \phi(r) + \phi(s) = (1,0) + (0,1) = (1,1) = (0,1) + (-1)(1,0) = \phi(sr^{-1})$.

Thus:

$\phi(1) = (0,0)$
$\phi(r) = (1,0)$
$\phi(r^2) = (0,0)$
$\phi(r^3) = (1,0)$
$\phi(s) = (0,1)$
$\phi(sr) = \phi(s) + \phi(r) = (1,0) + (0,1) = (1,1)$
$\phi(sr^2) = (0,1)$
$\phi(sr^3) = (1,1)$.

Note that $\text{ker }\phi = \{1,r^2\}$, and this is a normal subgroup of $D_4$.

The subgroups of $D_4$ are:

$D_4$
$\langle r \rangle = \{1,r,r^2,r^3\}$
$\langle r,s\rangle = \{1,r,s,sr\}$
$\langle r^2,sr\rangle = \{1,sr,r^2,sr^3\}$
$\langle r^2\rangle = \{1,r^2\}$
$\langle s\rangle = \{1,s\}$
$\langle sr\rangle = \{1,sr\}$
$\langle sr^2\rangle = \{1,sr^2\}$
$\langle sr^3\rangle = \{1,sr^3\}$
$\{1\}$.

Of these, the subgroups "over $\text{ker }\phi$" are:

$D_4$
$\langle r\rangle$
$\langle r^2,s\rangle$
$\langle r^2,sr\rangle$
$\langle r^2\rangle$

and $\phi$ induces the following bijection:

$D_4 \leftrightarrow \Bbb Z_2 \times \Bbb Z_2$
$\langle r\rangle \leftrightarrow \{(0,0),(1,0)\}$
$\langle r^2,s\rangle \leftrightarrow \{(0,0),(0,1)\}$
$\langle r^2,rs\rangle \leftrightarrow \{(0,0),(1,1)\}$
$\langle r^2\rangle \leftrightarrow \{(0,0)\}$

***************************

Why this is important:

This isomorphism theorem generalizes well to $R$-modules, and tells us how much of the internal structure of a module $M$ carries over to to a quotient module $M/N$. For example, If a subspace $U$ of a vector space $V$ properly contains the kernel (null space) of a linear transformation $T$, then $T(U)$ is a non-trivial subspace of $T(V)$. We can often use this to deduce when a system of linear equations has a non-trivial (non-zero) solution.

 

[blue_raver22]

's proposition, Proposition 5.2.3, states that in a simple group, every non-trivial normal subgroup is maximal. This means that there are no proper normal subgroups contained within the given normal subgroup.

To prove this proposition, we can use the fact that a simple group has no non-trivial normal subgroups. This means that any normal subgroup must either be the trivial subgroup {e} or the entire group itself.

Suppose we have a non-trivial normal subgroup N of a simple group G. Since N is normal, it must be invariant under conjugation by elements of G. This means that for any element g in G, the conjugate of N by g, denoted by gNg^-1, is also a subgroup of G.

Since N is non-trivial, it must contain at least one element other than the identity element e. Let's call this element n. Since G is simple, there are no proper normal subgroups, so n must generate the entire group G. This means that the subgroup generated by n, denoted <n>, is equal to G.

Now, consider the subgroup gNg^-1 for some arbitrary element g in G. Since n generates G, there must exist an element h in G such that g = nh. This means that gNg^-1 = (nh)N(nh)^-1 = n(hNh^-1)n^-1. Since N is normal, hNh^-1 is also a subgroup of G. But since n generates G, this means that n(hNh^-1)n^-1 = <n>. Therefore, gNg^-1 = <n>, which is equal to G.

This shows that for any element g in G, the conjugate of N by g is equal to G. This means that N is a maximal subgroup of G, since there are no proper subgroups contained within it. Therefore, Proposition 5.2.3 is proven.

As for Proposition 5.1.3, it states that in a simple group, every proper subgroup is contained in a maximal subgroup. This can be proven by using a similar argument as above, but instead of considering an arbitrary element g in G, we consider an arbitrary proper subgroup H of G. By the definition of a simple group, there are no proper normal subgroups, so H must be a maximal subgroup of G. Therefore, every proper subgroup in a simple group is contained in a maximal subgroup,