Answer to Question #179483 in Abstract Algebra for swarnajeet kumar

Question

a) Prove that every non-trivial subgroup of a cyclic group has finite index. Hence 

prove that (Q, +) is not cyclic. (7) 

 b) Let G be an infinite group such that for any non-trivial subgroup H of 

G, G : H < ∞. Then prove that 

 i) 

H ≤ G ⇒ H = {e} or H is infinite; 

 ii) If g ∈G, g ≠ e, then o(g) is infinite. (5) 

 c) Prove that a cyclic group with only one generator can have at most 2 elements. (3) 

Answer 1

a)

Given, G is a cylic group.

Therefore, G=<a>.

And H is subgroup of G.

H=<a^i>.

Index,

G/H={a^j+<a^i>}

If j>i then by division algorithm, j=ir+s

Then a^j+<a^i>=a^s+<a^i>

So, G/H={a^s+<a^i>, 0<=s<i}

O(G/H)= finite =index of H.

For( Q,+), choose H=<1/2>

Clearly o(G/H)= inifinte. 

So, G=(Q,+) is not cyclic.

b)

Given, G be an infinite group such that for any non-trivial subgroup H of 

G, G : H < ∞. 

i)

If H is finite then order of G will be finite by Lagrange's theorem. Which is contradiction.

Thus, H = {e} or H is infinite.

ii)

let g be any element of group G.

G=<g>

Since, H is infinte using H=<1>.

Therefore, o(g)=infinte.

c)

Let a be an arbitrary element of cyclic group G.

Since, G is cyclic.

Therefore, it generated by its elements say x i.e., G=⟨x⟩.

G=⟨x⟩, then there exists an integer k such that a=x^k=x^2q+r, where r∈{0,1},

r is the remainder of the euclidean division and q is the quotient.

Then a=(x²)^q.x^r=1^q.x^r=x^r, for r∈{0,1}. So, a can be 1 and x only.

Therefore, cyclic group G have atmost 2 elements.