question about Coset, Abstract Algebra?

Question

Suppose that G is a group with more than one element and G has no proper, nontrivial subgroups. Prove that |G| is prime. (Do not assumeat the outset that G is finite.) Show all your setp

Thank you

Answer 1

Let g be in G and g not = e, the identity,
if for every pair of distinct integers n and m, g^n not = g^m then G is an infinite group and {g^2n: n an integer} is a subgp. So it must be that
g^n = g^m for some distinct integers n and m and it follows G is a finite group (If you want to see the proof of this, write me at [e-mail].)
If H = {g^k: k = 0,1,2, . . . n} and H = G and |G| = pq, then g^q generates a subgp, so either |G| is a prime or H not = G.
But if not the latter, then H is a proper subgp of G. So, we're stuck with the order of G being a prime.

Answer 2

Let g (not 1) be in G.
Assume g^n=1 for some n (smallest possible), then the subgroup generated by g of G has order n. If the order of G is more than n, it is a proper subgroup. On the other hand, if the order of G is n but n is composite, n=pq, then the subgroup generated by g^p has order q and is a proper subgroup.
Now suppose g^n is not 1 for any n, then the subgroup generated by g^2 is a proper subgroup of G.
Thus, the |G| has to be prime.

Answer 3

John's answer is extremely exhaustive, mine is shorter: convinced that's an "iff" (if and on condition that) actuality. for coset multiplication to paintings, we choose each and every left coset of H to be a accurate coset (of an analogous representatives) as well. John's answer has the tedious data. (basically, if we pick (aH)(bH) = abH, then this suggests ahbh' = abh", so hbh' = bh" so hb = bh"h'^-a million. if H isn't regular, then there'll be some b for which Hb ? bH, so there'll be some h with hb no longer in bH, yet for coset multiplication to paintings, hb has to equivalent bh"h'^-a million it is obviously an aspect to bH). in spite of the indisputable fact that, you're in success: {?^0 = e,?^2} is definitely a common subgroup; enable's seem on the cosets: H = {e,?^2} ?H = {?,?^3} ?H = {?,??^2} ??H = {??,??^3} the right cosets are an analogous: H = {e.?^2} H? = {?,?^3} = ?H H? = {?,?^2?} = {?,??^2} = ?H because; ?^2? = ?(??) = ?(??^-a million) = ?(??^3) = (??)?^3 = (??^-a million)?^3 = ??^2 H?? = {??,?^2??} = {??,??^3} = ??H because: ?^2(??) = (??^2)? = (??^2)? (see above) = ??^3. so H is regular. now we are able to easily sq. each and every coset: (H)(H) = H (because H is a subgroup, it is continuously genuine). (?H)(?H) = ?^2H = H (because ?^2H = H, because ?^2 is in H). (?H)(?H) = ?^2H = H (because ?^2 = e). (??H)(??H) = (??)^2H = H (because (??)^2 = e: (??)^2 = (??)(??) = ?(??)? = ?(??^-a million)? = ?^2 = e). because H is the identity of G/H.....