# The Center of a Group

The **center of a group** $(G,*)$ consists of a subset Z(G) that includes every element $z \in G$ obeying the commutative property $g*z=z*g$ with any other element $g \in G$ under the group's operation. Essentially, it's defined as $$Z(G) = \{z \in G \ | \ z*g=g*z \ \ \forall \ g \in G\}$$.

Furthermore, the center, if it exists, is inherently a subgroup itself because it meets all the criteria of being a group.

Additionally, it's an Abelian subgroup, as it also fulfills the commutative property.

**Note**: In Abelian groups, the group's center coincides with the group itself. For instance, the center of the additive group of integers $(Z,+)$ regarding addition is the group Z(G)=$(Z,+)$ itself.

## A Practical Example

Let's consider the group G of permutations * of three objects {1,2,3}.

Within the group $(G,*)$, there are 6 possible permutations.

e=() - the identity

(12) - swaps 1 and 2

(13) - swaps 1 and 3

(23) - swaps 2 and 3

(123) - rotates 1 to 2, 2 to 3, 3 to 1

(132) - rotates 1 to 3, 3 to 2, 2 to 1

My task is to identify which permutations commute with all others in their composition.

In this case, the order of permutations affects the final outcome.

For example, composing (12) and (23) results in the permutation (132)

$$ (12)*(23)=(132) $$

**Explanation**: If the initial configuration was 123, the first permutation (23) from the right swaps the second and third position elements to achieve the intermediate result of 132. Then, the second permutation swaps the first and second position elements in the intermediate result, leading to the final result of 312. Therefore, in cyclic notation, the composition (12)*(23) is equivalent to the permutation (132) that rotates 1 to 3, 3 to 2, and 2 to 1.

**Note**: In compositions, perform the rightmost permutation first, followed by the left one. It's similar to composite functions f[g()] where the inner function g() is executed before the outer f().

However, composing (23) and (12) results in the permutation (123)

$$ (23)*(12)=(123) $$

**Explanation**: Starting with the initial configuration 123, the first permutation (12) swaps the first and second position elements, achieving an intermediate result of 213. The second permutation (23) then swaps the second and third position elements in the intermediate result, leading to the final result of 231. Thus, in cyclic notation, the composition (23)*(12) is equivalent to the permutation (123) that rotates 1 to 2, 2 to 3, and 3 to 1.

Therefore, the final outcome differs.

$$ (12)*(23) \ne (23) * (12) $$

This leads to the conclusion that the permutations (12) and (23) do not commute with the other permutations in the group.

Next, I'll construct a table of all possible compositions to analyze the rest.

e | (12) | (13) | (23) | (123) | (132) | |
---|---|---|---|---|---|---|

e | e | (12) | (13) | (23) | (123) | (132) |

(12) | (12) | e | (123) | (132) | (13) | (23) |

(13) | (13) | (132) | e | (123) | (23) | (12) |

(23) | (23) | (123) | (132) | e | (12) | (13) |

(123) | (123) | (23) | (12) | (13) | (132) | e |

(132) | (132) | (13) | (23) | (12) | e | (123) |

This visual illustrates how each pair of permutations merges to form a new permutation within the group.

- The first row and column deal with the identity element \(e\), which effectively leaves the paired permutation unchanged.
- The rest of the table reveals the result of combining two permutations.
Take, for example, merging \((12)\) with \((13)\) yields \((123)\), showcased at the cross-section of row \((12)\) and column \((13)\).

Wrapping up, the sole element of G that interacts seamlessly with all other elements, as illustrated by the composition table, is the identity element $ e $.

Thus, the nucleus of group G is:

$$ Z(G)=\{ e \} $$

Meaning, the essence of G is encapsulated in the identity permutation alone.

**Note**: In non-abelian (non-commutative) groups, such as this one, the center tends to be quite exclusive, often boiling down to the group's neutral element.

__Example 2__

This time, let's delve into the additive group \( (\mathbb{Z}, +) \), which frames integers in the light of addition.

The operation of adding integers is inherently commutative, translating to \(a + b = b + a\) for any integer pair \(a\) and \(b\).

This naturally implies that each element of \( \mathbb{Z} \) blends with every other, without exception.

Consequently, the heart of \( (\mathbb{Z}, +) \) is the whole of \( \mathbb{Z} \) itself, given that every element adheres to the commutativity criterion necessary for inclusion in the center.

$$ Z(\mathbb{Z}, +) = \mathbb{Z} $$

In essence, the center of any abelian (commutative) group aligns perfectly with the group as a whole.

## A Group's Center Is Its Subgroup

Within any group $ ( G, * ) $, the center $ Z(G) $ invariably forms a subgroup of G.

This principle is deeply rooted in the fundamental characteristics of the group's center.

**The Proof**

I'll check if the subgroup properties hold. Assuming the associative property is a given, since it's inherently satisfied by all elements of the group.

**Identity Element**

The identity element $ e = \{ \ \} $ doesn't alter the placement of elements. Therefore, it's always present in the center Z(G) and by definition commutes with all group elements. $$ e*g = g*e \ \ \ \forall \ g \ \in G $$**Closure Under Group Operation**

Any two elements \(a\) and \(b\) in the group's center \(Z(G)\), by definition, satisfy the commutative property, meaning for every element $ g \in G $, we have $ a*g = g*a $ and $ b*g = g*b $. I need to show that the product \(ab\) also commutes with every other group element \(g\) in \(G\), that is \(g(ab) = (ab)g\). Using the associative property I can write $$ g(ab) = (ga)b $$ Knowing \(a\) and \(b\) commute with \(g\) $$ g(ab) = (ga)b = (ag)b $$ Applying the associative property again $$ g(ab) = (ga)b = (ag)b = a(gb) $$ Since $ b \in Z(G) $ commutes with $ g $ $$ g(ab) = (ga)b = (ag)b = a(gb) = a(bg) $$ Finally, applying the associative property once more I get $$ g(ab) = (ga)b = (ag)b = a(gb) = a(bg) = (ab)g $$ Thus, the product $ ab $ belongs to the group's center $ Z(G) $ as it satisfies the commutative property. $$ g(ab) = (ab)g $$ This demonstrates that \(Z(G)\) is closed under the group operation.**Inverse Element for Every Member**

The last property to check is the existence of an inverse element for every member within the group's center $ Z(G) $. If an element \(a\) is in \(Z(G)\), then \(a\) commutes with every group element \(g\) in \(G\), meaning: $$ ag = ga $$ I need to prove that its inverse element \(a^{-1}\) also belongs to the group's center \(Z(G)\), which means \(a^{-1}\) must commute with every \(g \in G\).

Multiply both sides by $ a^{-1} $ on the left. $$ a^{-1} ag = a^{-1} ga $$ Knowing $ a^{-1}a=e $ where $ e $ is the identity element $$ eg = a^{-1} ga $$ Since $ eg=g $ $$ g = a^{-1} ga $$ Now, multiply both sides by $ a^{-1} $ but this time on the right. $$ g a^{-1} = a^{-1} gaa^{-1} $$ Knowing $ aa^{-1} = e $ $$ g a^{-1} = a^{-1} ge $$ Finally, since $ ge = g $ I conclude $$ g a^{-1} = a^{-1} g $$ This proves that the inverse element $ a^{-1} $ commutes with every element of the group. Hence, the inverse element belongs to the center $ a^{-1} \in Z(G) $.

In conclusion, the group's center meets all the group properties, making $ Z(G) $ a subgroup of $ G $.