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abstract algebra - If $G Z (G)$ is cyclic, then $G$ is abelian . . . The hint is actually the hardest part for me, as the quotient groups are somewhat abstract But once I have the hint, I can write: g, h ∈ G g, h ∈ G implies that g = xa1z1 g = x a 1 z 1 and h =xa2z2 h = x a 2 z 2, so
Prove that $Z(G)$ which is the center of $G$ is a subgroup of $G$ If I want to show that Z(G) Z (G) is a subgroup of G G that means I have to show that it is closed under group operation? Here is my attempt Let a, b a, b be elements in Z(G) Z (G) and x x be an element in G G Then ax = xa a x = x a which is under group multiplication commutative and under inverse (a−1)x = x(a−1) (a 1) x = x (a 1)
group theory - Prove that $|Z (G)|=p$ - Mathematics Stack Exchange While one can prove this using the oft-quoted result that if G Z(G) G Z (G) is cyclic then G G is abelian, this fact is not necessary to prove the result Assume, for the sake of contradiction, that the order of Z(G) Z (G) is exactly p2 p 2
Show that $Z(G) = \\{ x \\in G : gx = xg$ for all $g \\in G$} is a . . . Let G be a group and g ∈ G g ∈ G Show that Z(G) = {x ∈ G: gx = xg Z (G) = {x ∈ G: g x = x g for all g ∈ G g ∈ G} is a subgroup of G G This subgroup is called the center of G G I know that associativity can be inherited from G, but I'm not 100% sure on how to show it's closed, that an identity exists, and that an inverse exists
abstract algebra - Prove that $Z (G)$ is a subgroup of $C (a . . . If there exist an element g ∈ G g ∈ G such that C(g) = Z(G) C (g) = Z (G), prove that G G is a commutative group Deduce that in a non-commutative group G G, Z(G) Z (G) is a proper subgroup of C(g) C (g) for every g ∈ G g ∈ G
Is there a reason why $Z(G)$ is named the centre of a group? Finally, and this is mostly relevant for your question, Z(G) Z (G) turns out to be isomorphic to Θ ∩ Γ Θ ∩ Γ Then, in Sym(G) S y m (G) everything looks symmetric around the "center" Θ ∩ Γ Θ ∩ Γ: For clarity, I'm not saying this is really the reason why the center was historically named that way
$G$ is a group, prove: if $[G: Z(G)] lt;∞$, then $[G, G]$ is finite. You'll need to complete a few actions and gain 15 reputation points before being able to upvote Upvoting indicates when questions and answers are useful What's reputation and how do I get it? Instead, you can save this post to reference later
Groups - Prove that if $G Z(G)$ is cyclic then $G$ is abelian Prove that if G Z(G) G Z (G) is cyclic then G G is abelian Using this fact and G G is a nontrivial group of prime power order, deduce that a group of order p2 p 2 , p p prime, is abelian
abstract algebra - If $G H$ is cyclic, where $H$ is a subgroup of $Z (G . . . Second, in practice, it is the contrapositive of the theorem that is most often used - that is, if G G is non-Abelian, then G Z(G) G Z (G) is not cyclic For example, it follows immediately from the statement and Lagrange’s Theorem that a non-Abelian group of order pq p q, where p p and q q are prime, must have trivial center