Seeing if the multiplicative group $GF(2^{2016})$ has a subgroup of order 2016.

Question

How can I solve this.

I want to find out if the multiplicative geoup $GF(2^{2016})$ has a subgroup of order 2016. For this I need to check if 2016 divides $2^{2016}$.

Is there an easier way to see this without computing with such large numbers?

Thanks in advance

I have changed the title of this question to reduce confusion. Thank you all for your help.

Forgive me if I'm being obtuse (it's a bit late), but doesn't $3$ divide $2016$? How could $2016$ divide $2^k$ if $3$ doesn't divide $2^k$? — Michael L., Aug 12 '17 at 07:58
The multiplicative group in question has $2^{2016}-1$ elements (and is cyclic) — AdLibitum, Aug 12 '17 at 08:04
$2{2016}-1$ elements? Not the Euler totient function $=2^{2015}$? — Oscar Lanzi, Aug 12 '17 at 08:14
@OscarLanzi: he's not looking at the multiplicative group of ${\Bbb Z}_{2^{2016}}$. He's looking at the field with $2^{2016}$ elements (every element but $0$ is invertible) — AdLibitum, Aug 12 '17 at 08:37
@DariusTheGreat, See https://math.stackexchange.com/questions/1844558/how-to-find-last-two-digits-of-22016 and apply Carmichael function $$\lambda(63)=6$$ — lab bhattacharjee, Aug 12 '17 at 09:45

score 2 · Accepted Answer · answered Aug 12 '17 at 08:32

Certainly $2016$ does not divide $2^{2016}-1$.

If, more generally, you want to find the exact class of $2^{2016}$ in ${\Bbb Z}_{2016}$ since Euler's theorem does not apply you may proceed as follows.

First observe that $$ 2^{11}=2048\equiv2^5=32\bmod2016. $$ Then use this congruence to lower the exponent of $2$. Since $2016=11^3+5\cdot11^2+7\cdot11+3$ you get first $$ 2^{2016}=2^{11^3}\cdot(2^{11^2})^5\cdot(2^{11})^7\cdot2^3\equiv 2^{5^3}\cdot2^{5^3}\cdot2^{10}\cdot2^3\bmod2016 $$ (this is justified by $2^{ab}=(2^a)^b$) and then you proceed again "substituting" every $2^{11}$ in the congruence with a $2^5$ until the final exponent becomes feasable.

score 1 · Answer 2 · answered Aug 12 '17 at 08:12

$2016$ clearly doesn't divide $2^{2016}$ -- the only numbers that divide a power of $2$ are other (smaller) powers of $2$, and $2016$ is not a power of $2$ (it is strictly between $2^{10}=1024$ and $2^{11}=2048$).

However, as user438576 points oout, that is not the right question to ask.

You should be asking whether $2016$ divides $2^{2016}-1$, because that is the order of the multiplicative group of $GF(2^{2016})$ (which doesn't contain the zero of the field).

When does an even number divide an odd number?

score 0 · Answer 3 · answered Aug 12 '17 at 10:09

I see now. First of all the multiplicative group $GF(2^{2016})$ has $2^{2016} - 1$ elements, since zero is not in it. It is now easy to see that 2016 can't divide $2^{2016}$, hence there's no subgroup of order 2016 for this group.

I have changed the title of this question to reduce confusion. Thank you all for your help.

Seeing if the multiplicative group $GF(2^{2016})$ has a subgroup of order 2016.

3 Answers3