Abstract algebra: if G is cyclic, prove that G/N is cyclic
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Abstract algebra: if G is cyclic, prove that G/N is cyclic

[From: ] [author: ] [Date: 11-12-21] [Hit: ]
where N is any subgroup of G.Thanks for the help!-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.......
If G is a cyclic group, prove that G/N is cyclic, where N is any subgroup of G.

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!
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