Question

14. Suppose that G is a group in which {1} and G are the only subgroups. Show that G is finite and, in fact, is cyclic of order 1 or a prime.

Answer 1

0 Suppose that is a Coop Paving no non toivio Subindoupie G har have suboroup {0} and is it holy only. then we prove that must be finite and order of is poime G. Suppose G has infinite order. then up Cane find af staze Let H=<a>, then hiq Cyclic subbroup But G has no non- trivial SubGroups. thoa - H= G - G=<a> of G and H 7503 Consider now o(a)s 91-1 OBSENTA the SubGroup K=< 97 > Now a4 cozy because it ae<a27 then a=0 25 for some imoges => 027-1.4 = meaining thereby that oca) is finito which is not tove. thue 94<927 Again 2+ $83, because then ad-f would again mean that Ola) is finite (<2) thue <a2 is a HON - trivial suboroup of a which is not Posible. Hence o(G) cannot be infinite. So oo is finite. ☺ orm) is poime. 8 413 i) = $13 then dC)=1 & Satiaty Given Condition. ... Let How Gl ia finite Composite, HD 0(G)=n=r9 where <r, san.

Since n71, 7 etata Congides go. Case-I qraf => 0(a)=8 lot oca)=k. => 1 <ksoan (>1 aa que) let Ha $ 9, 94, 93 -- Ke3. then his non-empty dinite Suban set of G. and it is closed Unded multiplication. → H iG 9 Subcrgoup oj . Since OCH=K4n which is contradiction. G has no have non trivial Subbroup. Case I gote then since (ar)- 980-an -oola- e. orgo ) = 5 let 0(90)=4 Hheo 1 <$<S<n if we take k= $98,928 -- 928 e3 then k isa non-empty finite Subset of a closed under multiplication & in Subhroup of G OCK) <n. which again Contradiction. S has no have non trivial SubGroup. 1.P our adsumption is woong ola) is finite Composite ooder → 0(6)=213 or P-(Poime).