Given set $S=\{1,2,3,5,7,8,9\}$ $(\{1,2,3,5,7,8,9\},\bigotimes_{10})$ is not a group.
Option by option:
- It is not closed
$(2\times 3) mod \;10=6 \notin S$
$(2\times 5) mod \;10=0 \notin S$
$(2\times 2) mod \;10=4 \notin S$
So we can observe $\{0,4,6\}$ these are not in $S$. This group is not closed as they not belong to our base set.
- $2$ doesn’t have an inverse
$(2\times 1) mod \;10=2 \in S$
$(2\times 2) mod \;10=4 \notin S$
$(2\times 3) mod \;10=6 \notin S$
$(2\times 5) mod \;10=0 \notin S$
$(2\times 7) mod \;10=4 \notin S$
$(2\times 8) mod \;10=6 \notin S$
$(2\times 9) mod\;10=8 \in S$
So, here is no identity element. Thus the inverse of $2$ not even exist.
- $3$ doesn't have an inverse
No. $3$ has an inverse, which is $7$
$(3\times 7)mod\;10=1;\;\;(7\times 3)mod\;10=1\;\;\therefore7=3^{-1}$
- $8$ doesn’t have an inverse
$(8\times 1) mod \;10=8 \in S$
$(8\times 2) mod \;10=6 \notin S$
$(8\times 3) mod \;10=4 \notin S$
$(8\times 5) mod \;10=0 \notin S$
$(8\times 7) mod \;10=6 \notin S$
$(8\times 8) mod \;10=4 \notin S$
$(8\times 9) mod\;10=2 \in S$
So, here is no identity element. Thus the inverse of $8$ not even exist.
$\color{DarkGreen} Only\;C\;is\;false$