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Let $S= \{0, 1, 2, 3, 4, 5, 6, 7\}$ and $⊗$ denote multiplication modulo $8,$ that is, $x ⊗ y= (xy) \mod 8$

  1. Prove that $( \{ 0, 1\}, ⊗)$ is not a group.
  2. Write three distinct groups $(G, ⊗)$ where $G ⊂ S$ and $G$ has $2$ elements.
asked in Set Theory & Algebra by Veteran (52.1k points)
edited by | 816 views

2 Answers

+14 votes
Best answer

$\begin{array}{c|cc}
⊗&0&1\\ \hline
0&0&0\\
1&0&1
\end{array}$

a. $1$ is the identity element. Inverse does not exist for zero. So, it is not a group.

b. $\begin{array}{c|cc}
⊗&1&3\\ \hline
1&1&3\\
3&3&1
\end{array}\qquad\begin{array}{c|cc}
⊗&1&5\\\hline1&1&5\\
5&5&1
\end{array}\qquad\begin{array}{c|cc}
⊗&1&7\\\hline1&1&7\\
​7&7&1
\end{array}$

answered by Loyal (5.6k points)
edited by
0
Then how can A be a group?
0
Sorry I can't get your question.Actual question is to prove A is not a group.I also proved it.Is there is any mistake?
+1
Sorry. i misread the question and didn't see the A, B separation in answer.
+1
it's ok :)
0 votes

({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.

answered by Active (3.4k points)
0
{1,3,5,7} with order 4

with order 2 no further subgroup exist
0
Inverse of 7 is 7

The other two groups are {1,3}, {1,5}

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