group theory

[logistic_guy]

Prove that $\displaystyle Z_i(G)$ is a characteristic subgroup of $\displaystyle G$ for all $\displaystyle i$.

 

[khansaheb]

Prove that $\displaystyle Z_i(G)$ is a characteristic subgroup of $\displaystyle G$ for all $\displaystyle i$.

show us your effort/s to solve this problem.

 

[logistic_guy]

show us your effort/s to solve this problem.

Here is my approach to solve this problem. For now, I will just solve it for the case when $\displaystyle i = 1$.

I will introduce three elements $\displaystyle \varphi, a \ \text{and} \ b$.
I will let
$\displaystyle \varphi \in \text{Aut(G)}$
$\displaystyle a \in Z_1(G)$
and
$\displaystyle b \in G$

$\displaystyle \varphi \in \text{Aut(G)}$ means that $\displaystyle \varphi$ is an automorphism which is bijective and has the following properties:
$\displaystyle \varphi(xy) = \varphi(x)\varphi(y)$ for any $\displaystyle x, y \in G$
$\displaystyle \varphi(\varphi^{-1}(x)) = x$ for all $\displaystyle x \in G$
$\displaystyle xy = yx$ for any $\displaystyle y \in G$

The main idea is to show $\displaystyle \varphi(a) \in Z_1(G)$ where $\displaystyle Z_1(G) = Z(G)$ is the center of $\displaystyle G$.

Now I want to verify that $\displaystyle \varphi(a)b = b\varphi(a)$. By applying the properties of $\displaystyle \varphi$ on the combination $\displaystyle \varphi(a)b$, I get:

$\displaystyle \varphi(a)b = \varphi(a)\varphi(\varphi^{-1}(b)) = \varphi(a\varphi^{-1}(b)) = \varphi(\varphi^{-1}(b)a) = \varphi(\varphi^{-1}(b))\varphi(a) = b\varphi(a)$

I have just shown that for every $\displaystyle a \in Z_1(G), \varphi(a) \in Z_1(G)$ which means that $\displaystyle Z_1(G)$ is a characteristic subgroup of $\displaystyle G$.

Does my proof miss any important details, or is it complete?

 

[fresh_42]

Prove that $\displaystyle Z_i(G)$ is a characteristic subgroup of $\displaystyle G$ for all $\displaystyle i$.

What is a characteristic subgroup and what does the index $i$ stand for?

 

[logistic_guy]

What is a characteristic subgroup?

I know that the center $\displaystyle Z(G) := \{x \in G \ | \ xg = gx \ \forall \ g \in G\}$ is a subgroup of $\displaystyle G$

$\displaystyle Z(G)$ is a characteristic subgroup of $\displaystyle G$ if $\displaystyle \varphi(Z(G)) \subset Z(G)$ for all automorphisms $\displaystyle \varphi$ of $\displaystyle G$

what does the index $i$ stand for?

It is the $\displaystyle i^{\text{th}}$ center. It is defined like this:

$\displaystyle Z_{0}(G) = G$, the group itself
$\displaystyle Z_{1}(G) = Z(Z_{0}(G)) = Z(G)$, the center
$\displaystyle Z_{2}(G) = Z(Z_1(G)) = Z(Z(G))$, the center of the center
$\displaystyle \cdots$
$\displaystyle Z_{i+1}(G) = Z(Z_{i}(G))$, the center of the $\displaystyle i^{\text{th}}$ center

 

[fresh_42]

The center of a group is abelian, so the center of the center is the entire center, and $Z_k(G)=Z(G)$ for all $k>0.$
$$\varphi(z)\cdot g=\varphi\left(z\cdot \varphi^{-1}(g)\right)=\varphi \left(\varphi^{-1}(g)\cdot z\right)=g\cdot \varphi(z)$$This shows that the center is invariant under automorphisms, and the group itself is, of course, also invariant.

This is the formal answer to your question using your definition. However, I suspect that there is something wrong since $$\ldots = Z_3(G)=Z_2(G)=Z_1(G)=Z(G) \trianglelefteq G$$is a pretty meaningless chain. Maybe you meant the central series of a group.

 

[logistic_guy]

Maybe you meant the central series of a group.

I can't tell what's the difference between them. I am just mimicking the subject for now. To fully understand the meaning of the notations takes time.

Thanks professor for sharing some of your great knowledge.

 

[tomwebb64]

Explore Your Desires with Honesty and Consent.

 

[khansaheb]

To fully understand the meaning of the notations takes time.

So does reviewing "basic" pre-requisite subjects - in addition it is embarrassing ......