# FP3 Groups Table HelpWatch

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#1
Hi,

So I need help with q6 on this paper.

http://pmt.physicsandmathstutor.com/...0FP3%20OCR.pdf

Using Lagranges theorem I was able to find the order of the elements. But can someone help me find the number of elements which have that order. I created a table but I wasn't able to find the number of elements.

Thanks
0
1 year ago
#2
(Original post by ChemBoy1)
Hi,

So I need help with q6 on this paper.

http://pmt.physicsandmathstutor.com/...0FP3%20OCR.pdf

Using Lagranges theorem I was able to find the order of the elements. But can someone help me find the number of elements which have that order. I created a table but I wasn't able to find the number of elements.

Thanks
I'm assuming you've done 6 i)

For ii) I think the crux is that if G is not a cyclic group, and has order 4, then it must not contain any element with order 4. If it contains that element (call it b), then b^2 will be in, likewise, b^3, and likewise b^4 = e. Therefore it is a cyclic group, which is a contradiction.
This narrows down the options for G such that if G is a subgroup of Q, then it must only contain the elements with orders 1 , 2 and 3. (There are 4 of these, so all 4 of these are in the group)

I don't really fully understand the question though if I'm honest. Are G and H definitely subgroups, because it never mentions that, it just says that they're non-cyclic groups
0
1 year ago
#3
(Original post by have)
I don't really fully understand the question though if I'm honest. Are G and H definitely subgroups, because it never mentions that, it just says that they're non-cyclic groups
G and H are general groups of the order given. They are not necessarily subgroups of Q (so you can't use the table of orders for Q) - in fact the latter part of the question is to show that they can't be.
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1 year ago
#4
(Original post by ChemBoy1)
...
Although you cannot use the table for Q, have makes an important point - you cannot have an element with the same order as the group, else the group would be cyclic.

You should also know the number of elements of order 1. Think on it, if necessary.

That's sufficient to polish off G.

H requires more work. Consider: Must there be an element of order 3?
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#5
Hi thanks for this, it really helps! I'm just a bit confused on one thing. For a group containing 4 elements (lets say where Z = 4 except for 0 1, 2, 3 and 4) what would be the order of 4 be??
(Original post by ghostwalker)
Although you cannot use the table for Q, have makes an important point - you cannot have an element with the same order as the group, else the group would be cyclic.

You should also know the number of elements of order 1. Think on it, if necessary.

That's sufficient to polish off G.

H requires more work. Consider: Must there be an element of order 3?
0
1 year ago
#6
(Original post by ChemBoy1)
Hi thanks for this, it really helps! I'm just a bit confused on one thing. For a group containing 4 elements (lets say where Z = 4 except for 0 1, 2, 3 and 4) what would be the order of 4 be??
If it contains 4 elements, then the standard additive group is {0,1,2,3} with the group operation being addition modulo 4.

4, as such, is not an element of the group.

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