- Center (group theory)
In

abstract algebra , the**center**of a group "G" is the set "Z"("G") of all elements in "G" which commute with all the elements of "G". That is,:$Z(G)\; =\; \{z\; in\; G\; |\; gz\; =\; zg\; ;forall,g\; in\; G\}$.

Note that "Z"("G") is a

subgroup of "G", because

# "Z"("G") contains "e", theidentity element of "G", because "eg" = "g" = "ge" for all "g" ∈ G by definition of "e", so by definition of "Z"("G"), "e" ∈ "Z"("G");

# If "x" and "y" are in "Z"("G"), then ("xy")"g" = "x"("yg") = "x"("gy") = ("xg")"y" = ("gx")"y" = "g"("xy") for each "g" ∈ "G", and so "xy" is in "Z"("G") as well (i.e., "Z"("G") exhibits closure);

# If "x" is in "Z"("G"), then "gx" = "xg", and multiplying twice, once on the left and once on the right, by "x"^{−1}, gives "x"^{−1}"g" = "gx"^{−1}— so "x"^{−1}∈ "Z"("G").Moreover, "Z"("G") is an abelian subgroup of "G", a

normal subgroup of "G", and even a strictlycharacteristic subgroup of "G", but not always fully characteristic.The center of "G" is all of "G"

if and only if "G" is an abelian group. At the other extreme, a group is said to be**centerless**if "Z"("G") is trivial, i.e. consists only of the identity element.**Conjugation**Consider the map "f": "G" → Aut("G") from "G" to the

automorphism group of "G" defined by "f"("g") =_{"g"}, where_{"g"}is the automorphism of "G" defined by :$phi\_g(h)\; =\; ghg^\{-1\}\; ,$.This is agroup homomorphism , and its kernel is precisely the center of "G", and its image is called theinner automorphism group of "G", denoted Inn("G"). By thefirst isomorphism theorem we get:$G/Z(G)cong\; m\{Inn\}(G).$Thecokernel of this map is the group $operatorname\{Out\}(G)$ ofouter automorphism s, and these form theexact sequence ::$1\; o\; Z(G)\; o\; G\; o\; operatorname\{Aut\}(G)\; o\; operatorname\{Out\}(G)\; o\; 1.$**Examples*** The center of the group $mbox\{GL\}\_n(F)$ of "n"-by-"n" invertible matrices over the field $F$ is the collection of scalar matrices $\{\; sI\_n\; |\; s\; in\; Fsetminus\{0\}\; \}$.

* The center of theorthogonal group $O(n,\; F)$ is $\{\; I\_n,-I\_n\; \}$.

* The center of thequaternion group $Q\; =\; \{1,\; -1,\; i,\; -i,\; j,\; -j,\; k,\; -k\}$ is $\{1,\; -1\}$.

* The center of the multiplicative group of non-zeroquaternion s is the multiplicative group of non-zero real numbers.

* Using theclass equation one can prove that the center of any non-trivial finitep-group is non-trivial.

* Non-abeliansimple group s are centerless.

* If thequotient group $G/Z(G)$ is cyclic, G is abelian.**Higher centers**Quotienting out by the center of a group yields a sequence of groups called the

::$G\_0\; =\; G\; o\; G\_1\; =\; G\_0/Z(G\_0)\; o\; G\_2\; =\; G\_1/Z(G\_1)\; o\; cdots$The kernel of the map $G\; o\; G\_i$ is theupper central series **"i"th center**of "G" (**second center**,**third center**, etc.), and is denoted $Z^i(G)$.Following this definition, one can define the 0th center of a group to be the identity subgroup. This can be continued totransfinite ordinals bytransfinite induction ; the union of all the higher centers is called the. [hypercenter *This union will include transfinite terms if the UCS does not stabilize at a finite stage.*]The ascending chain of subgroups:$1\; leq\; Z(G)\; leq\; Z^2(G)\; leq\; cdots$stabilizes at "i" (equivalently, $Z^i(G)\; =\; Z^\{i+1\}(G)$)

if and only if $G\_i$ is centerless.**Examples*** For a centerless group, all higher centers are zero, which is the case $Z^0(G)=Z^1(G)$ of stabilization.

* ByGrün's lemma , the quotient of aperfect group by its center is centerless, hence all higher centers equal the center. This is a case of stabilization at $Z^1(G)=Z^2(G)$.**References****ee also***

center (algebra)

*centralizer and normalizer

*conjugacy class .

*Wikimedia Foundation.
2010.*