If G is a cyclic group, prove that G/N is cyclic, where N is any subgroup of G.
Thanks for the help!
Thanks for the help!
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Suppose that G = for some g in G.
Since N is a subgroup of G (which is cyclic), N is also cyclic.
==> N = for some non-negative integer k.
So, any element in G/N may be written as
(g^j) N for some non-negative integer j.
By the division algorithm, j = qk + r for some integers q, r with r in {0, 1, 2, ..., k-1}.
Hence, (g^j)N
= (g^r * (g^k)^q) N, via division algorithm
= g^r N, since (g^k)^q is in N
= (gN)^r, by coset multiplication.
So, we see that = G/N; hence G/N is cyclic.
I hope this helps!
Since N is a subgroup of G (which is cyclic), N is also cyclic.
==> N =
So, any element in G/N may be written as
(g^j) N for some non-negative integer j.
By the division algorithm, j = qk + r for some integers q, r with r in {0, 1, 2, ..., k-1}.
Hence, (g^j)N
= (g^r * (g^k)^q) N, via division algorithm
= g^r N, since (g^k)^q is in N
= (gN)^r, by coset multiplication.
So, we see that
I hope this helps!