Originally Posted by adam_leeds the inverse of -1 is -1 and is in the set 1 times -1 = -1 which is in the set which shows closure but i thought you said 1 couldnt be in the subset? No - the subset cannot be precisely $\displaystyle \{1\}$ if it is trivial, but 1 must be in it.
Follow Math Help Forum on Facebook and Google+
Originally Posted by Swlabr No - the subset cannot be precisely $\displaystyle \{1\}$ if it is trivial, but 1 must be in it. so the non trivial subgroups are {1,-1} {1,i} {1, -i} {1,-1,i} {1,-1,-i}
Originally Posted by adam_leeds so the non trivial subgroups are {1,-1} {1,i} {1, -i} {1,-1,i} {1,-1,-i} No. Remember what the theorem you are trying to show says. Which of your sets don't adhere to what it says? The problem with some of your given sets is closure: $\displaystyle a, b \in H \Rightarrow ab \in H$.
Originally Posted by Swlabr No. Remember what the theorem you are trying to show says. Which of your sets don't adhere to what it says? The problem with some of your given sets is closure: $\displaystyle a, b \in H \Rightarrow ab \in H$. bottom two because -1 * 1 * i = -i which is not in the subgroup -1 * 1 * -i = i which is not in the subgroup
Originally Posted by adam_leeds bottom two because -1 * 1 * i = -i which is not in the subgroup -1 * 1 * -i = i which is not in the subgroup Yes, but you just multiply two elements together and not all of them. The product of any two elements in the set must be in the set.
Originally Posted by Swlabr Yes, but you just multiply two elements together and not all of them. The product of any two elements in the set must be in the set. i love you, cheers for your help
View Tag Cloud