536 views
2 votes
2 votes

                   Is ( {0},* ) a group ?   ( where * stands for multiplication operation ).

Please provide explanations with your answer 

1 Answer

Best answer
6 votes
6 votes

Group: For any algebraic structure to be a group, that has to satisfy the Closure, Associatively, Identity and Inverse properties. 

Closure

For all $a, b \in G$, the result of the operation, $a * b$, is also in $G$.

In above example, Since there is one element, hence $ a = b = 0 $, and $ a * b = 0 * 0 = 0 \in G $

Hence Closure satisfy.

Associative

For all $a, b, c \in G$, $ (a * b) * c = a * (b * c) $.

For above example, $ a = b = c = 0 $

Hence $ (a * b) * c = a * (b * c) $

 $ \implies (0 * 0) * 0 = 0 * (0 * 0) \implies 0 = 0 $

Hence Associatively satisfied.

Identity element

There exists an element $e \in G$ such that, for every element $a \in G$, the equation $ e * a = a * e = a $ holds. Such an element is unique, and thus one speaks of the identity element.

For above example $ a = e = 0 $ 

Hence $ e * a = a * e \implies 0 * 0 = 0 * 0 \implies  0 = a $

Hence $ e = 0 $ is the identity element.  

Inverse element

For each $a \in G$, there exists an element $b \in G$, commonly denoted $a^{−1}$, such that $a * b = b * a = e $, where $e$ is the identity element.

For your example, The inverse element is $0$. Because when you multiply $0$ with $0$ then you will get $0$, which is also an identity element of the structure. 

selected by

Related questions

1 votes
1 votes
2 answers
1
Vicky rix asked Mar 25, 2017
869 views
"the union of two sub-groups neednot be a sub-group".can some-body prove without using counter example ...
1 votes
1 votes
1 answer
3
ankitgupta.1729 asked May 1, 2018
1,228 views
Prove that :-Every infinite cyclic group is isomorphic to the infinite cyclic group of integers under addition.
0 votes
0 votes
1 answer
4
Deepesh Pai asked Mar 6, 2018
390 views
Let $G$ be a finite group with sub group $H$ & $K$ such that $|H|=7$ and $|K|=31$ then find $| H ⋂ K|$