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show $b_1b_2b_3\cdots b_{\phi(m)} \equiv 1 \pmod{m}$ or $b_1b_2b_3\cdots b_{\phi(m)} \equiv -1 \pmod m$

where $b_1 < b_2 < b_3<\cdots< b_{\phi(m)}$ are the integers between $1$ and $m$ that are relatively prime to $m$.

Seems like a simple enough exercise, but i am running into a dead end on how to start. Suggestions on how to begin? I'd prefer not to get a full solution but some breadcrumbs to lead me in the right directon.

I tnought of am idea similar to what was used in the proof of Fermat's little theorem, but that wouldn't work.

edit: no group theory solutions, any hints if they can be restricted only to number theory.

Michael Hardy
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D.C. the III
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  • Do you mean relatively prime to $m$? – B. Freitas Feb 03 '16 at 18:01
  • yes you were right @B.Freitas – D.C. the III Feb 03 '16 at 18:08
  • Possible duplicate of http://math.stackexchange.com/questions/53026/prove-that-a-1a-2-cdots-a-n2-e-in-a-finite-abelian-group. – lhf Feb 03 '16 at 18:08
  • I posted this answer. This explains how to find a multiplicative inverse modulo a prime number. But the only fact about that prime number that was needed was that it is coprime to the number whose inverse we seek. Thus each of the numbers $b_1,\ldots,b_{\varphi(m)}$ has a multiplicative inverse that is one of $b_1,\ldots,b_{\varphi(m)}$. If $b$ is one of those, and $b\ne b^{-1}$, then${},\ldots\ldots\qquad$ – Michael Hardy Feb 03 '16 at 18:26
  • $\ldots$ the product $b_1\cdots b_{\varphi(m)}$ includes as a factor the product $bb^{-1}=1$. After those cancellations are done, one is left with the product of those that are their own inverses. I'm not sure yet how to proceed after that. $\qquad$ – Michael Hardy Feb 03 '16 at 18:26
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    @MichaelHardy: Let $m\gt 2$. We have $s$ is self-inverse iff $m-s$ is, and they are distinct. The product of $s$ and $m-s$ is congruent to $-1$. If There is an odd number of such pairs of self-inverses, the full product is congruent to $-1$. If there is an even number then it is congruent to $1$. – André Nicolas Feb 03 '16 at 18:33
  • Multiplication modulo $m$ on $b_1,..,b_{\psi (m)}$ is a finite commutative group. When you multiply all the members of a finite commutatative group ,you get....? A special case, when $p$ is prime, is $(p-1)!\equiv -1\pmod p.$ – DanielWainfleet Feb 04 '16 at 04:06

3 Answers3

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Idea: If $i\ne j$ and $b_ib_j\equiv 1\pmod{m}$, call $b_i$ and $b_j$ buddies. The product of all people who are members of a buddy pair is congruent to $1$.

Call the rest of the $b_k$ solitary. Note that if $m\gt 2$ then solitaries $s$ also can be grouped into pairs, namely $s$ and $m-s$.

André Nicolas
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  • should i have known from before that "product of all people who are members of a buddy pair is congruent to 1" ? I am trying to remember where I habe seen or established this result – D.C. the III Feb 03 '16 at 18:24
  • The result does not need proof. Suppose $a$ and $b$ are buddies, also $c$ and $d$, also $e$ and $f$. Then the product $abcdef$ is $(ab)(cd)(ef)$ and each of the three terms is congruent to $1$. Some in general. Now you need to write down the analogous argument for solitaries. – André Nicolas Feb 03 '16 at 18:28
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    I am feeling really uncomfortable about how you were able to state that $b_ib_j \equiv 1 \pmod{m}$ from our set of elements. Then I figure that same idea is used for the solitaries, but the idea with buddies I am having issue wrapping my head around it. – D.C. the III Feb 03 '16 at 18:42
  • Every $b$ relatively prime to $m$ has an inverse $c$ modulo $m$, which is also relatively prime to $m$. If $c\not \equiv b\pmod{m}$, then $b$ and $c$ are what I called buddies. You probably have seen a proof that every $b$ relatively prime to $m$ has an inverse. The fancy way is to use the Bezout identity, there exist $s$ and $t$ such that $bs+mt=1$, so $bs\equiv 1\pmod{m}$. There are less fancy ways, that I prefer. For example consider $bb_1, bb_2, \dots, bb_{\varphi(m)}$. It is not hard to show that these are pairwise incongruent, so one of the products must be congruent to $1$. – André Nicolas Feb 03 '16 at 19:01
  • ok after fiddling with it and thinking about the definitions for awhile, I could understand that since $b_i$ are relatively prime to $m$ then each of them has a multiplicative inverse, as seen by Berzout's thm. So I can rearrange my $b_i's$ to match up with their inverses: $b_ib_j \equiv 1 \pmod{m}$ ,but how would I show for -1? How do the self inverse elements as you call them work? – D.C. the III Feb 03 '16 at 20:48
  • thinking about it more wouldn't it depend on whther $\phi(m)$ is odd or even? If it is not even then I woukd be left with a leftover $b_i$. – D.C. the III Feb 03 '16 at 20:50
  • I dealt with these, calling them solitary. If $s$ is solitary, so is $m-s$. Their product is congruent to $-s^2$, so congruent to $-1$. The product of these $-1$'s is congruent to $-1$ or $1$. – André Nicolas Feb 03 '16 at 20:52
  • is this because $$s(m-s) = ms - s^2$$ and if $m$ divides $ms - s^2$ their remainder is $s^2$ . Where m is the m from (mod m). Then that would mean that the sign of my product of $b_i's$ depends on how many solitaries I have. what is the relationship between s and m-s? – D.C. the III Feb 03 '16 at 21:06
  • Yes, whether if is $-1$ or $1$ depends on whether the number of solitaries is congruent to $2$ or $0$ modulo $4$. (For $m\gt 2$ the number of solitaries is even, since $s$ is solitary iff $m-s$ is.) – André Nicolas Feb 03 '16 at 21:22
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Hint: We will bring this down with some group theory. Let $\mathbb{Z}^*_m$ stand for the multiplicative group of numbers mod m. This is a group, so each element has an inverse; that is, for each $n\in\mathbb{Z}^*_m$ there is another $n^{-1}\in\mathbb{Z}^*_m$ such that $n*n^{-1}=1$. See where I am going?

Jsevillamol
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The statement in the title of the question is true iff $U(m)$ is not cyclic. In particular, it is false for $m$ prime.

The correct statement folows from Wilson's theorem for finite Abelian groups:

The product of all elements in a finite Abelian group is either $1$ or the element of order $2$ if there is only one such element.

lhf
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