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.
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.
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