Algebra-1-G-is-a-group-and-o-G-p-2-prove-that-G-is-an-abelian-group-hint-p-is-prime-number-

Question Number 192399 by mnjuly1970 last updated on 17/May/23

A l g e b r a (1)

G, i s a g r o u p a n d o (G) = p^{2} .

p r o v e t h a t G i s a n a b e l i a n g r o u p .

h i n t : (p i s p r i m e n u m b e r)

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Commented by aleks041103 last updated on 21/May/23

o (G) i s t h e o r d e r o f t h e g r o u p, r i g h t ?

I . e . t h e n u m b e r o f e l e m e n t s i n G

Answered by aleks041103 last updated on 21/May/23

{l e t}^{'} s u s e t h e c e n t e r o f G, i . e .

Z (G) = Z = {z \in G ∣ g z = z g, \forall g \in G}

i t i s k n o w n t h a t Z ⊴ G .

b y {l a g r a n g e}^{'} s t h e o r e m :

o (Z) o (G / Z) = o (G) = p^{2}

\Rightarrow o (Z) = 1, p, p^{2}

\underset{―}{1 s t c a s e :}

i f o (Z) = p^{2} = o (G), t h e n G = Z \Rightarrow G i s a b e l i a n .

\underset{―}{2 n d c a s e :}

i f o (Z) = p, t h e n o (G / Z) = p .

e v e r y g r o u p o f p r i m e o r d e r i s c y c l i c .

\Rightarrow \exists a \notin Z, s . t . G / Z = {a^{k} Z ∣ k = 0, 1, \dots, p - 1}

\Rightarrow \forall g \in G, \exists z \in Z, \exists k \in Z_{p} : g = a^{k} z

\Rightarrow l e t g_{1}, g_{2} \in G \Rightarrow g_{1} = a^{j} z_{1} a n d g_{2} = a^{k} z_{2}

\Rightarrow g_{1} g_{2} = a^{j} z_{1} a^{k} z_{2}

s i n c e z_{1, 2} \in Z t h e y c o m m u t e w i t h e v e r y t h i n g

\Rightarrow g_{1} g_{2} = a^{j} z_{1} a^{k} z_{2} = a^{k} z_{2} a^{j} z_{1} = g_{2} g_{1}

\Rightarrow G i s a b e l i a n a n d Z \equiv G .

\underset{―}{3 r d c a s e :}

i f o (Z) = 1, Z = {e} .

S i n c e G i s a p - g r o u p (o (G) = p^{n}), i t i s k n o w n

t h a t t h e c e n t e r Z (G) i s n o n t r i v i a l, i . e .

Z (G) \neq {e} .

C o n c l u s i o n :

G i s a b e l i a n .