Algebra (1 ) G, is a group and o(G ) = p^( 2) . prove that G is an abelian group. hint: ( p is prime number ) −−−−−−−−−−−−−


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

$A l g e b r a (1)$ $G, i s a g r o u p a n d o (G) = p^{2} .$ $p r o v e t h a t G i s a n a b e l i a n g r o u p .$ $h i n t : (p i s p r i m e n u m b e r)$ $- - - - - - - - - - - - -$
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
	$let′s use the center of G, i.e. Z(G)=Z={z∈G∣gz=zg, ∀g∈G} it is known that Z⊴G. by lagrange′s theorem: o(Z)o(G/Z)=o(G)=p^2 ⇒o(Z)=1,p,p^2 1st case: if o(Z)=p^2 =o(G), then G=Z⇒G is abelian. 2nd case: if o(Z)=p, then o(G/Z)=p. every group of prime order is cyclic. ⇒∃a∉Z, s.t. G/Z={a^k Z∣k=0,1,...,p−1} ⇒∀g∈G,∃z∈Z, ∃k∈Z_p : g=a^k z ⇒let g_1 ,g_2 ∈G⇒g_1 =a^j z_1 and g_2 =a^k z_2 ⇒g_1 g_2 =a^j z_1 a^k z_2 since z_(1,2) ∈Z they commute with everything ⇒g_1 g_2 =a^j z_1 a^k z_2 =a^k z_2 a^j z_1 =g_2 g_1 ⇒G is abelian and Z≡G. 3rd case: if o(Z)=1, Z={e}. Since G is a p−group(o(G)=p^n ), it is known that the center Z(G) is nontrivial, i.e. Z(G)≠{e}. Conclusion: G is abelian.$
	${l e t}^{'} s u s e t h e c e n t e r o f G, i . e .$ $Z (G) = Z = {z \in G ∣ g z = z g, \forall g \in G}$ $i t i s k n o w n t h a t Z ⊴ G .$ $b y {l a g r a n g e}^{'} s t h e o r e m :$ $o (Z) o (G / Z) = o (G) = p^{2}$ $\Rightarrow o (Z) = 1, p, p^{2}$ $\underset{―}{1 s t c a s e :}$ $i f o (Z) = p^{2} = o (G), t h e n G = Z \Rightarrow G i s a b e l i a n .$ $\underset{―}{2 n d c a s e :}$ $i f o (Z) = p, t h e n o (G / Z) = p .$ $e v e r y g r o u p o f p r i m e o r d e r i s c y c l i c .$ $\Rightarrow \exists a \notin Z, s . t . G / Z = {a^{k} Z ∣ k = 0, 1, . . ., p - 1}$ $\Rightarrow \forall g \in G, \exists z \in Z, \exists k \in Z_{p} : g = a^{k} z$ $\Rightarrow l e t g_{1}, g_{2} \in G \Rightarrow g_{1} = a^{j} z_{1} a n d g_{2} = a^{k} z_{2}$ $\Rightarrow g_{1} g_{2} = a^{j} z_{1} a^{k} z_{2}$ $s i n c e z_{1, 2} \in Z t h e y c o m m u t e w i t h e v e r y t h i n g$ $\Rightarrow g_{1} g_{2} = a^{j} z_{1} a^{k} z_{2} = a^{k} z_{2} a^{j} z_{1} = g_{2} g_{1}$ $\Rightarrow G i s a b e l i a n a n d Z \equiv G .$ $\underset{―}{3 r d c a s e :}$ $i f o (Z) = 1, Z = {e} .$ $S i n c e G i s a p - g r o u p (o (G) = p^{n}), i t i s k n o w n$ $t h a t t h e c e n t e r Z (G) i s n o n t r i v i a l, i . e .$ $Z (G) \neq {e} .$ $C o n c l u s i o n :$ $G i s a b e l i a n .$
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