Question

5-Determinc an upper bound for the size of the automarphism group for the follouing grop. Lt helps to fivst determine elements here are oF each orden mavny how

answer this question and the final answer must be ?

Answer 1

Z15 = {1, 2,4,7,8, 11, 13, 14}
. Among these elements, 2,7,8 and 13 are of order 4 and 4,11 and 14 are of order 2. Now suppose
0 :215 +Z15
is an automorphism. Then
o(1) = 1
. Also 2 and $\phi (2)$ must be of same order. So, $\phi (2)$ can be one of 2,7,8 and 13. Now,
((2)) = (22) ف = (4)
and for all the four choices of $\phi(2)$ , we have
(6(2)=4
and hence
4 = (4) في
. Now once $\phi(2)$ is chosen, $\phi(8)$ gets fixed as
o(8) = (23) = ($(2)
. Now 11 is of order 2 , so $\phi(11)$ must be of order 2. Hence, $\phi(11)$ can be either 11 or 14. [Note that 4 is also or order 2 but we already have a pre-image of 4 which is 4 itself]. Now, once $\phi(11)$ is chosen, $\phi(14)$ is also fixed as $\phi(14)$ is also of order 2, it must be one of 11 and 14, so if
o(11) = 11
then
(14) = 14
and if
o(11) = 14
then
(14) = 11
. Now once $\phi(11)$ and $\phi(2)$ are both chosen, $\phi(7)$ and $\phi(13)$ get fixed as
(11) (2) = (7) في
and
(14) (2) = (13) في
. Since $\phi(2)$ has 4 choices and $\phi(11)$ has 2 choices, in total we can choose them both in 8 ways. Each of these choices can give a different automorphism and any automorphism on $\mathbb{Z}_{15}^*$ will satisfy one of these 8 choices. So, 8 is an upper bound for the size of the automorphism group of $\mathbb{Z}_{15}^*$ .