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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
Algebra (1 ) G, is a group and o(G ) = p^( 2) . prove that G is an abelian group. hint: ( p is prime number ) −−−−−−−−−−−−−
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Algebra}\:\left(\mathrm{1}\:\right) \\ $$$$\:\:{G},\:{is}\:{a}\:{group}\:\:{and}\:\:\:{o}\left({G}\:\right)\:=\:{p}^{\:\mathrm{2}} \:. \\ $$$$\:\:\:{prove}\:{that}\:{G}\:{is}\:{an}\:{abelian}\:{group}. \\ $$$$\:\:\:{hint}:\:\:\left(\:{p}\:{is}\:{prime}\:{number}\:\:\right) \\ $$$$\:\:\:\:\:−−−−−−−−−−−−− \\ $$
Commented by aleks041103 last updated on 21/May/23
o(G) is the order of the group, right? I.e. the number of elements in G
$${o}\left({G}\right)\:{is}\:{the}\:{order}\:{of}\:{the}\:{group},\:{right}? \\ $$$${I}.{e}.\:{the}\:{number}\:{of}\:{elements}\:{in}\:{G} \\ $$
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.
$${let}'{s}\:{use}\:{the}\:{center}\:{of}\:{G},\:{i}.{e}.\: \\ $$$${Z}\left({G}\right)={Z}=\left\{{z}\in{G}\mid{gz}={zg},\:\forall{g}\in{G}\right\} \\ $$$${it}\:{is}\:{known}\:{that}\:{Z}\trianglelefteq{G}. \\ $$$${by}\:{lagrange}'{s}\:{theorem}: \\ $$$${o}\left({Z}\right){o}\left({G}/{Z}\right)={o}\left({G}\right)={p}^{\mathrm{2}} \\ $$$$\Rightarrow{o}\left({Z}\right)=\mathrm{1},{p},{p}^{\mathrm{2}} \\ $$$$ \\ $$$$\underline{\mathrm{1}{st}\:{case}:} \\ $$$${if}\:{o}\left({Z}\right)={p}^{\mathrm{2}} ={o}\left({G}\right),\:{then}\:{G}={Z}\Rightarrow{G}\:{is}\:{abelian}. \\ $$$$ \\ $$$$\underline{\mathrm{2}{nd}\:{case}:} \\ $$$${if}\:{o}\left({Z}\right)={p},\:{then}\:{o}\left({G}/{Z}\right)={p}. \\ $$$${every}\:{group}\:{of}\:{prime}\:{order}\:{is}\:{cyclic}. \\ $$$$\Rightarrow\exists{a}\notin{Z},\:{s}.{t}.\:{G}/{Z}=\left\{{a}^{{k}} {Z}\mid{k}=\mathrm{0},\mathrm{1},…,{p}−\mathrm{1}\right\} \\ $$$$\Rightarrow\forall{g}\in{G},\exists{z}\in{Z},\:\exists{k}\in\mathbb{Z}_{{p}} \::\:{g}={a}^{{k}} {z} \\ $$$$\Rightarrow{let}\:{g}_{\mathrm{1}} ,{g}_{\mathrm{2}} \in{G}\Rightarrow{g}_{\mathrm{1}} ={a}^{{j}} {z}_{\mathrm{1}} \:{and}\:{g}_{\mathrm{2}} ={a}^{{k}} {z}_{\mathrm{2}} \\ $$$$\Rightarrow{g}_{\mathrm{1}} {g}_{\mathrm{2}} ={a}^{{j}} {z}_{\mathrm{1}} {a}^{{k}} {z}_{\mathrm{2}} \\ $$$${since}\:{z}_{\mathrm{1},\mathrm{2}} \in{Z}\:{they}\:{commute}\:{with}\:{everything} \\ $$$$\Rightarrow{g}_{\mathrm{1}} {g}_{\mathrm{2}} ={a}^{{j}} {z}_{\mathrm{1}} {a}^{{k}} {z}_{\mathrm{2}} ={a}^{{k}} {z}_{\mathrm{2}} {a}^{{j}} {z}_{\mathrm{1}} ={g}_{\mathrm{2}} {g}_{\mathrm{1}} \\ $$$$\Rightarrow{G}\:{is}\:{abelian}\:{and}\:{Z}\equiv{G}. \\ $$$$ \\ $$$$\underline{\mathrm{3}{rd}\:{case}:} \\ $$$${if}\:{o}\left({Z}\right)=\mathrm{1},\:{Z}=\left\{{e}\right\}. \\ $$$${Since}\:{G}\:{is}\:{a}\:{p}−{group}\left({o}\left({G}\right)={p}^{{n}} \right),\:{it}\:{is}\:{known} \\ $$$${that}\:{the}\:{center}\:{Z}\left({G}\right)\:{is}\:{nontrivial},\:{i}.{e}. \\ $$$${Z}\left({G}\right)\neq\left\{{e}\right\}. \\ $$$$ \\ $$$$\boldsymbol{{Conclusion}}: \\ $$$${G}\:{is}\:{abelian}. \\ $$

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