Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the question:
compute $\phi(24)$ for each element Z/24 decide whether the element is a unit or a zero divisor, if the element is a unit divisor give its order and find its inverse.

Ive worked out $\phi(24)=8$
and the unit divisors to be ${1,5,7,11,13,17,19,23}$
however when I came to working out the order I got them all to be 2?

share|cite|improve this question
Presumably, you mean all but one of them has order 2. – Gerry Myerson Nov 28 '12 at 12:16
yupp bar the number 1, could you confirm ive answered it correctly or have i gone terribly wrong? – jill Nov 28 '12 at 12:18
@jill You can write $(\mathbb{Z}/24)^{\times} \cong (\mathbb{Z}/8)^{\times} \times (\mathbb{Z}/3)^{\times} \cong \mathbb{Z}/2 \times \mathbb{Z}/2 \times \mathbb{Z}/2$ in which it's easy to see that all elements have order at most $2$ – Cocopuffs Nov 28 '12 at 12:27

If you look at the numbers mod 6 (the prime divisors of 24 being 2 and 3) you eliminate those equivalent to 0,2,3,4 - so you have left those equivalent to 1 and 5 (i.e. $6n \pm 1$). You could confirm your answer by looking at what happens to the numbers $(6n \pm 1)^2$ when they are taken mod 24.

share|cite|improve this answer

$(2a+1)^2=4a^2+4a+1=8\frac{a(a+1)}2+1\equiv \pmod 8\implies c^2\equiv1\pmod 8$ if $(c,8)=1$

Now, $(3b\pm1)^2\equiv 1\pmod 3\implies c^2\equiv1\pmod 3$ if $(c,3)=1$

If $(c,8)=1$ and $(c,3)=1$ i.e., if $(c,24)=1, lcm(3,8)\mid (c^2-1)\implies c^2\equiv1\pmod{24}$

So, $ord_{24}c\mid 2$

But clearly, $ord_{24}c\ne1$ for $c>1\implies ord_{24}c=2$ for $c>1$ and $(c,24)=1$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.