Question

The set $\{1,2,3,5,7,8,9\}$ under multiplication modulo 10 is not a group. Given below are four possible reasons. Which one of them is false ?

• A. It is not closed
• B. 2 does not have an inverse
• C. 3 does not have an inverse
• D. 8 does not have an inverse

Answer 1

The correct option is C 3 does not have an inverse

Let $A=\{1,2,3,5,7,8,9\}$
Construct the table for any $x,yϵA$ such that
$x\ast y=(x.y)\text{mod}10$
$\begin{array}{cccccccc}\ast & 1& 2& 3& 5& 7& 8& 9\\ 1& 1& 2& 3& 5& 7& 8& 9\\ 2& 2& 4& 6& 0& 4& 6& 8\\ 3& 3& 6& 9& 5& 1& 4& 7\\ 5& 5& 0& 5& 5& 5& 0& 5\\ 7& 7& 4& 1& 5& 9& 6& 3\\ 8& 8& 6& 4& 0& 6& 4& 2\\ 9& 9& 8& 7& 5& 3& 2& 1\end{array}$
We know that $0\text{/}ϵA$. So it is not closed. Therefore, (a) is true.
The identity element = 1
$\therefore \left({2.2}^{-1}\right)\text{mod}10=1$
From the table we see that $2^{-1}$ does not exist.
Since, $(3,7)\text{mod}10=1$
$\therefore 7$ is the inverse of $3$ and $7ϵA$
$\therefore$ (c) is false.
(d) is true since 8 does not have inverse.