1

The Question:

Let n be a positive integer and let $G_n = \left\{[a] ∈ \mathbb{Z}_n ; \text{gcd}(a,n) = 1\right\}$ be the group of invertible elements in (Zn,·), where ”·” represents the product (mod n).

Prove that $(G_{125},·)$ is a group with 100 elements. Use Lagrange’s theorem to find all possible sizes of subgroups of $G_{125}$. Hence prove that [2] is a generator for $(G_{125}, ·)$. (You may use without checking the following identities (mod 125): $2^{10} ≡24,2^{20} ≡76,2^{25} ≡57$)

My Attempt: We can see that $\left|G_{125}\right| = 100$ by using the fact that only multiples of 5 in $\mathbb{Z}_{125}$ are not in $G_{125}$, of which there are 25.

By Lagrange, the order of a subgroup $<d>$ of $G$ divides the order of $G$. So the set of all possible sizes of subgroups in $G_{125}$ $:= \left\{a ; \text{gcd}\left(a,100\right) = a, a\in\mathbb{Z}\right\} = \left\{1,2,4,5,10,20,25,50,100\right\}$

At this point I'm stuck. The logic I tried using is that if [2] (mod 125) is a generator of $G_{125}$ then the cyclic subgroup $<[2]>$ should have the same order as $G_{125}$, that is, $2^{\text{ord}\left(G_{125}\right)}≡1$ (mod 125).

It's clear from above that the order of this subgroup can only be one of the numbers from the set of possible sizes of subgroups. It's not going to be any of the first 7 elements (simple calculations and the hint given in the question show this).

So I'm left with 50 and 100. How do I show that the order of $<[2]>$ is 100 and not 50? ($2^{50}$ and $2^{100}$ are huge unusable numbers.)

Or am I using the wrong method of proving that [2] generates $G_{125}$?

yerman
  • 455
  • 1
    Hint: You don't need to compute $2^{50}.$ You only need to compute $[2]^{50}$ – Brian Moehring Aug 13 '19 at 17:22
  • 1
    Remember you are are working $\pmod {125}$, $2^{50}$ is quite manageable. We have $2^7\equiv 3$ for instance. Work from that. – lulu Aug 13 '19 at 17:22
  • So would factorising 50 help? – yerman Aug 13 '19 at 17:29
  • 1
    here is a good general result, though this particular case can be handled on its own. – lulu Aug 13 '19 at 17:29
  • "Prove that (G125,⋅) is a group with 100 elements." Does this mean we have to prove $\phi(125) = 100$? Or that ${[a]:\gcd(a,n)},\cdot$ forms a group? The way the question is stated it would seem we can take both of those as given. – fleablood Aug 13 '19 at 18:04
  • It's pretty clear that $G_n$ must be a subset of $Z_n$ $\forall n$. I think more relevant to the question was the understanding of the fact that $5\mathbb{Z}$ does not intersect $G_{125}$. – yerman Aug 13 '19 at 18:16
  • "At this point I'm stuck. The logic I tried using is that if [2] (mod 125) is a generator of G125 then the cyclic subgroup <[2]> should have the same order as G125, that is, 2ord(G125)≡1 (mod 125)." Right! The order of $2$ must divide $100$ and the order of $2$ is not $10$, $20$ or $25$. It can't be $2$ or $5$ because $2^{10}\not \equiv 1$. That leave $ord(2)=50$ or $ord(2)=100$. There may be some clever way to show $2^{50}\not \equiv 1$ but $2^{50}\equiv 57^2\equiv(225 +7)^2\equiv 42525 + 4257 + 49\equiv 20125 + 2525+325+49\equiv 20125+5125+ 325+225-1\equiv -\pmod {125}$. – fleablood Aug 13 '19 at 18:17
  • 1
    @rory_c $G_n$ is not a subgroup of $\mathbb{Z}_n.$ $G_n$ is a group and $G_n \subseteq \mathbb{Z}_n$ as sets, but $( \mathbb{Z}_n, \cdot)$ isn't a group – Brian Moehring Aug 13 '19 at 18:19
  • *subset. Wasn't thinking. – yerman Aug 13 '19 at 18:20
  • I'm just saying it's odd to say: "prove" $G_{125}$ has $100$ elements. By definition $G_{125}$ are the set of equivalence classes where $a$ are relatively prime to $125$ and that, by definition, is $\phi(125)$ which simple calculation is $100$. I just found the wording rather bizarre is all. – fleablood Aug 13 '19 at 18:22
  • "250 and 2100 are huge unusable numbers.)" Not really. You were given $2^{25} \equiv 57$ so $2^{50} \equiv 57^2$ and that isn't huge. – fleablood Aug 13 '19 at 18:24
  • looks anything greater than $1$ is huuuege :P – AgentS Aug 13 '19 at 18:30
  • Yeah lol kick the college student cause he forgot how mod works after only learning it last week lol good laugh. – yerman Aug 13 '19 at 18:35
  • @rory_c jk :D If you just want to check $2^{50}\equiv 1$ you can easily do that from noticing just the last digit of $57^2$. – AgentS Aug 13 '19 at 19:08

2 Answers2

1

Hint if $\, 2^{\large n}\equiv 1\pmod{\!125}$ then ditto $\!\bmod 5\,$ so $\,4\mid n,\,$ by $\,2\,$ has order $4$. But $\,4\nmid 50$.

Or $\bmod 5^{\large 3}\!:\ 2^{\large 50} = (2^{\large 10})^{\large 5}\equiv (-1\!+\!25)^{\large 5}\equiv (-1)^{\large 5}\, $ via Binomial Theorem (other terms $\equiv 0)$

Bill Dubuque
  • 272,048
0

Hint:
$2^{50} \equiv 1 \implies 2^{25}\equiv \pm 1 $

In general if $x^2\equiv 1\pmod{p^e}$, where $p$ is an odd prime,
then $x\equiv \pm 1 \pmod{p^e}$

AgentS
  • 12,195
  • 1
    So is this a contradiction? Couldn't you say $2^{100} \equiv 1 \implies 2^{50} \equiv \pm 1$? – yerman Aug 13 '19 at 17:29
  • 1
    Yes, since $2^{25}$ is $57$, we can say $2^{50}$ cannot be $1$. This means order of $2$ is $\gt 50$ – AgentS Aug 13 '19 at 17:31
  • 1
    That makes sense, thanks. Does the above imply that $2^{50}$ must be -1, then? Or am I understanding it incorrectly? – yerman Aug 13 '19 at 17:33
  • 1
    That's right. $2^{100} \equiv 1$ and $2^{25} \not\equiv 1$ forces $2^{50}\equiv -1$ – AgentS Aug 13 '19 at 17:34
  • 2
    Note, however, that $x^2 \equiv 1 \pmod m$ does not imply $x \equiv \pm 1 \pmod m$ for general $m,$ so one would need to prove it applies to $m=125$ to use it. – Brian Moehring Aug 13 '19 at 17:35
  • 1
    @BrianMoehring I agree.. $x^2-1 = (x-1)(x+1)$. If $x-1$ and $x+1$ share a common factor, then $x\equiv \pm 1$ thingy breaks – AgentS Aug 13 '19 at 17:38
  • 1
    How would you go about proving it? I'm not sure how to do it rigorously without showing it for every equivalence relation. – yerman Aug 13 '19 at 17:39
  • 1
    @rory_c I'm a bit rusty with these as I did these long ago. Gimme some time.. I'll get back :) – AgentS Aug 13 '19 at 17:40
  • 1
    @rory_c $x^2 - 1 \equiv 0 \pmod{125} \implies 125 \mid (x+1)(x-1)$. Since $x=57$ here and neither $57+1$ or $57-1$ has a factor $5$, we're good – AgentS Aug 13 '19 at 17:51
  • 1
    Since $125$ is the power of a prime, the proof is not hard. Suppose $$(x-1)(x+1) \equiv 0 \pmod{5^3}.$$ Then one of the following holds $$x-1 \equiv 0 \pmod{5^3} \ x+1 \equiv 0 \pmod{5^3} \ x-1 \equiv x+1 \equiv 0 \pmod{5}.$$ From here, just show the last case isn't possible with $p=5.$ – Brian Moehring Aug 13 '19 at 17:51
  • 1
    I'm a little leery of saying $2^{50}\equiv 1\pmod{125}\implies 2^{25}\equiv \pm 1$. $2^{50}\equiv 1\implies 2^{50}-1=(2^{25}-1)(2^{25}+1)\equiv 0\pmod{125}$ but that doesn't imply $2^{25}\pm 1\equiv 0$. It could mean $2^{25}\pm1 \equiv 5$ and $2^{25}\mp 1\equiv 25$. This would mean $5\mp 1 \equiv 25\pm 1$ which is obviously false so it does imply that but we shouldn't encourage students to make assumptions about division or roots of modulo arithmetic. – fleablood Aug 13 '19 at 18:33
  • @fleablood totally agree! Highly likely it may lead students to think it works for any $\mod n$ – AgentS Aug 13 '19 at 18:37
  • @fleablood alternatively you could arrive at that from noticing $5 \nmid 2 = (a^{m}+1) - (a^{m}-1)$ – AgentS Aug 13 '19 at 18:48
  • Along the same lines as the concern @fleablood mentioned, I believe the following is true (as a consequence of the chinese remainder theorem): $$\forall x \in \mathbb{Z} \left(x^2 \equiv 1 \pmod{m} \implies x \equiv \pm 1 \pmod{m} \right) \ \iff \ m = p^e \text{ or } m = 2p^e \text{ for some odd prime } p$$ which gives you an idea of how special this property is. – Brian Moehring Aug 13 '19 at 18:59
  • There's been enough counterexamples in my life to make me weary. Using Brian Moehrings example for $n=20$ then $3^4\equiv 1\pmod {20}$ but $2^2 \equiv 9\equiv -11\pmod{20}$. – fleablood Aug 13 '19 at 19:33
  • @fleablood Don't understand 20 is not a prime power – AgentS Aug 13 '19 at 19:36
  • That's why it's a counterexample. – fleablood Aug 13 '19 at 20:17
  • Still don't understand – AgentS Aug 13 '19 at 20:17