1.2k views

The set $\{1, 2, 4, 7, 8, 11, 13, 14\}$ is a group under multiplication modulo $15$. The inverses of $4$ and $7$ are respectively:

1. $3$ and $13$
2. $2$ and $11$
3. $4$ and $13$
4. $8$ and $14$

edited | 1.2k views

Option C.

Identity element here is $1.$
$4 * 4 \mod 15 = 1$
$7 * 13 \mod 15 = 1.$
by Active (3.3k points)
selected by
+10

It is same as finding modular inverse, which is also last step in RSA.

You may likely to know actual procedure for even big numbers-

https://goo.gl/ZCza9r

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How the identity element is 1?
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An Identity element is an element which leaves other elements unchanged when combined with them.

e.g a # e = a  Here e is an identity element, # is any random operator.

For multiplication,1 multiplying with any other number gives that number itself.

e.g,   5 x 1 = 5

Hence 1 is an identity element with respect to multiplication.
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$\\ 4d\equiv 1(mod15)\\ GCD(4,15)=1\ How?\\ 15=4*3+3\\ 4=3*1+\mathbf{1}\\ Now\ to\ find\ inverse\ of\ 4\ backtrack\ euclidean\ algorithm\\ 1=4-3*1\\ 1=4-1(15-4*3)\\ 1= 4.4-15.1\\ Taking\ (Mod 15) \ on\ both\ sides\\ 1mod(15)=4.4\\ Hence\ inverse\ of\ 4\ is\ 4$

## The correct answer is,(c) 4 and 13

by Loyal (7.6k points)

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