# is the set a group wrt to multiplication

• Mar 29th 2009, 06:15 PM
yvonnehr
is the set a group wrt to multiplication
Is the set a group wrt multiplication. If not, what all conditions does it fail? If so, what's the order.

The set {[1], [3]} ⊆ Z(sub)8.

The book says it's a group of order 2.

I made a Cayley table but have no clue how to check associativity, and inverse. I know it's closed by theorem.
But how do I check associativity? I'm confused.

If {[1]} = {[1], [2], [3], [4], [5], [6], [7]}, and
{[3]} = {[3], [6], [1], [4], [7], [2], [5]}, then is
{[1], [3]} = [1]}? I don't know the relevance of this. It's just something that I am confused about.

Identity is [1], that was clear in the table.

Inverse: Rule [a] [b] = [1]. How do I check this?

Thanks for all your help. This is blowing my mind!(Headbang)
• Mar 29th 2009, 06:20 PM
NonCommAlg
Quote:

Originally Posted by yvonnehr
Is the set a group wrt multiplication. If not, what all conditions does it fail? If so, what's the order.

The set {[1], [3]} ⊆ Z(sub)8.

The book says it's a group of order 2.

I made a Cayley table but have no clue how to check associativity, and inverse. I know it's closed by theorem.
But how do I check associativity? I'm confused.

If {[1]} = {[1], [2], [3], [4], [5], [6], [7]}, and
{[3]} = {[3], [6], [1], [4], [7], [2], [5]}, then is
{[1], [3]} = [1]}? I don't know the relevance of this. It's just something that I am confused about.

Identity is [1], that was clear in the table.

Inverse: Rule [a] [b] = [1]. How do I check this?

Thanks for all your help. This is blowing my mind!(Headbang)

{[1], [3]} with multiplication is a subgroup of order 2. the identity element is [1] and note that [3].[3] = [9] = [1].
• Mar 29th 2009, 06:32 PM
yvonnehr
huh?
Thanks, but I don't know that means.