Math Help - Lagranges theorem

1. 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 $\{1\}$ if it is trivial, but 1 must be in it.

2. Originally Posted by Swlabr
No - the subset cannot be precisely $\{1\}$ 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}

3. 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: $a, b \in H \Rightarrow ab \in H$.

4. 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: $a, b \in H \Rightarrow ab \in H$.
bottom two because -1 * 1 * i = -i which is not in the subgroup

-1 * 1 * -i = i which is not in the subgroup

5. 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.

6. 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

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