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I've been investigating as part of a project the structure of integers modulo $n$ ($\mathbb{Z}$/n$\mathbb{Z}$) under multiplication. One aspect I'm looking at is, for any natural number $k$, finding a value of $n$ such that $\mathbb{Z}/n\mathbb{Z}$ admits a group of order $k$, where $1$ is not the identity element of this group.

E.g. For $k=5$, $\{2,4,8,16,32\}$ is a group in $\mathbb{Z}/62\mathbb{Z}$.

My conjecture is that, for any $k>1$, the set $A=\{2, 4, 8,...,2^k\}$ is a group of order $k$ in $\mathbb{Z}/n\mathbb{Z}$, where $n=2^{k+1}-2$. I have shown that $A$ satisfies the various group axioms, other than it being closed under multiplication.

That is, given any $x,y\in A,\space xy\in A$. Which is equivalent to saying $xy\equiv 2^i\pmod{n}$ for $1\leq i\leq k$. I've spent a fair amount of time trying to prove this and I'm not sure if I'm unable to either because of the fact I'm not very strong at dealing with congruences, or because it is actually untrue. I have an intuition in my head of why it should(?) be true, but if anyone could offer me any assistance with this, a nudge in the right direction, I'd appreciate it greatly.

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2 Answers 2

up vote 2 down vote accepted

It is enough (why?) to prove that $2\times 2^k$ lies in your set, but this is obvious because $2\times 2^k= 2^{k+1} \equiv 2 \mod 2^{k+1}-2$.

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In fact you are viewing the subgroup of powers of $2$ in $\mathbb{Z}/(2^k-1)\mathbb{Z}$ inside $\mathbb{Z}/(2^{k+1}-2)\mathbb{Z}$ (via multiplication by 2). –  Quimey Apr 11 '13 at 16:06
    
I think you want $\pmod {2^{k+1}-2}$ –  Ross Millikan Apr 11 '13 at 16:19
1  
@Quimey: Thanks a bunch! It's obvious now. –  WakeUpDonnie Apr 11 '13 at 16:24

If $l+m<k+1$, then $2^l2^m=2^{l+m}$ is clearly in your set.

If $l+m>k$, then

$$2^{l+m}\equiv2^{l+m}-(2^{k+1}-2)2^{l+m-k-1}=2^{l+m-k}.$$

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Thanks for that. Very straightforward. –  WakeUpDonnie Apr 11 '13 at 16:25

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