Let $S= \{0, 1, 2, 3, 4, 5, 6, 7\}$ and $⊗$ denote multiplication modulo $8,$ that is, $x ⊗ y= (xy) \mod 8$
$\begin{array}{c|cc} ⊗&0&1\\ \hline 0&0&0\\ 1&0&1 \end{array}$
a. $1$ is the identity element. Inverse does not exist for zero. So, it is not a group.
b. $\begin{array}{c|cc} ⊗&1&3\\ \hline 1&1&3\\ 3&3&1 \end{array}\qquad\begin{array}{c|cc} ⊗&1&5\\\hline1&1&5\\ 5&5&1 \end{array}\qquad\begin{array}{c|cc} ⊗&1&7\\\hline1&1&7\\ 7&7&1 \end{array}$
({0,1},⊗) is not a group because : (0e)mod8=0 is haing e=0 as identity but (1e)mod8=1 should have 1 as identity so no single identity is there.
({1,7}, ⊗ ) :
1. closed
2. associative
3. Identity = 1
4. inerse: 1^{-1}=7 , 7^{-1}=1
can someone suggest 2 more subgroups.
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