Every subgroup of $(\mathbb {Z_n},+)$ is closed under multiplication
1,105
$(\Bbb Z_n,+)$ is a cyclic group, and any subgroup $G$ is likewise cyclic. This means, in particular, that if $a,b \in G$, that:
$a = kg = g + g +\cdots + g\ (k$ times)
$b = mg = g + g + \cdots + g\ (m$ times)
for some integers $k,m$ and a generator $g$ of $G$.
Now compute $ab$ (mod $n$) -can you think of a way to somehow use the distributive law for integers, and then reduce mod $n$?
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Bosnia
Updated on March 02, 2020Comments
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Bosnia over 3 years
I am stuck in this proof that every subgroup of $(\mathbb {Z_n},+)$ is also a subring. which requires me to prove it is closed under multiplication. I have to show if $a,b \in G<\mathbb {Z_n}$ , then $ab \in G<\mathbb {Z_n}$
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Michael Burr over 8 yearsI suggest starting with describing what the subgroups of $\mathbb{Z}_n$ are. This problem will be much easier if you describe what $G$ looks like.
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John McGee over 8 yearsInteger multiplication can be modeled as repeated addition, i.e. 5a = a+a+a+a+a.
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Carl Mummert over 8 yearsGetting stuck is not too hard - what did you try? I would suggest induction on $b$, and prove that $G$ is closed under multiplication by every natural number.
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Bosnia over 8 years@JohnMcGee thanks. So $ab \in G$ because $ab = b+b+...a times \in G$
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Bosnia over 8 years$ab=(km)g$, correct?
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David Wheeler over 8 yearsYes, but you should be able to prove this (induction will work, but arguing from integers and reducing mod $n$ is "faster").
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abel over 8 yearsis $g = gcd(a,b)?$