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# Algebra tricky question watch

1. Very confused on parts 1 and 2 of this, not sure of the proof at all,
And how do I go abouts part 3?
Thankyou for any help in advance
2. (Original post by maths10101)

Very confused on parts 1 and 2 of this, not sure of the proof at all,
And how do I go abouts part 3?
Thankyou for any help in advance
Can you post a proper picture - this is cut off a little. What are your thoughts on it?
3. (Original post by Zacken)
Can you post a proper picture - this is cut off a little. What are your thoughts on it?

Sorry about that and ignore part c, I've figured fhat one out..
For part a, I'm completely clueless of how to carry it out
For part b, would you find the orders of all elements of A^6? And use this to show that g is 1? There must be a quicker way of doing this as this method is quite time consuming
4. (Original post by maths10101)

Sorry about that and ignore part c, I've figured fhat one out..
For part a, I'm completely clueless of how to carry it out
For part b, would you find the orders of all elements of A^6? And use this to show that g is 1? There must be a quicker way of doing this as this method is quite time consuming
For a) and b) surely you could just apply Lagrange?
As for b) I wouldn't want to find all the elements of A6, there are 360 of them.
5. (Original post by joostan)
For a) and b) surely you could just apply Lagrange?
As for b) I wouldn't want to find all the elements of A6, there are 360 of them.
So for a) i've seen that since the cardinality of the subgroup divides the cardinality of the group (S-12), it must be a subgroup by lagrance theorem,,,,does this seem okay?
and for part b....i'm not sure how to prove that
6. (Original post by maths10101)
So for a) i've seen that since the cardinality of the subgroup divides the cardinality of the group (S-12), it must be a subgroup by lagrance theorem,,,,does this seem okay?
and for part b....i'm not sure how to prove that
I don't follow what you're saying for a), Lagrange's Theorem states that the order of a subgroup must divide the order of the group.
What is the order of ? What can you then say about the order of the subgroups of ?

As for (b) how many elements of are there?
Now if you suppose that an element of order 7 exists, what can you say about the set ?
7. (Original post by joostan)
I don't follow what you're saying for a), Lagrange's Theorem states that the order of a subgroup must divide the order of the group.
What is the order of ? What can you then say about the order of the subgroups of ?

As for (b) how many elements of are there?
Now if you suppose that an element of order 7 exists, what can you say about the set ?
The order of S_12 is 479001600
and i'm not finding anything about the order of the subgroups online?
8. (Original post by maths10101)
The order of S_12 is 479001600
and i'm not finding anything about the order of the subgroups online?
It has order 12! - does 17 divide that? Prove that it doesn't.
9. (Original post by Zacken)
It has order 12! - does 17 divide that? Prove that it doesn't.
Ahhh brilliant! Any ideas on part 2 mate?
10. (Original post by maths10101)
The order of S_12 is 479001600
and i'm not finding anything about the order of the subgroups online?
You know that the order of is .
You want to show that there is no subgroup of order .
If a group of order has a subgroup of order , then we must have by Lagrange.
Apply this to this case.

Again, for the second question, the order of an element must divide the order of the group by Lagrange.
Thus if then what are the possible orders of ?
11. For Part 3) use the method of successive squaring.
12. (Original post by fishyfishy1999)
For Part 3) use the method of successive squaring.
He's already done part 3. :-)
13. (Original post by Zacken)
He's already done part 3. :-)
Oops. The answer is 10 for what it is worth.
14. (Original post by fishyfishy1999)
Oops. The answer is 10 for what it is worth.
Given that he's already done it, mind posting an outline of how you do it? I'm interested in knowing and I haven't heard of your method before. If it's not too much trouble, that is.
15. (Original post by Zacken)
Given that he's already done it, mind posting an outline of how you do it? I'm interested in knowing and I haven't heard of your method before. If it's not too much trouble, that is.

It's a pain to write down in this box. Just Google Successive Squaring Algorithm for computing a^k mod m.

Quite straightforward.

End up with 18 14 18 14 18 14 18 18 7 10 7 7 10 14 7 10 14 18 10 as the non-zero remainders that are multiplied together to give 439205094557429760000 mod 31 which is 10.
16. (Original post by fishyfishy1999)
It's a pain to write down in this box. Just Google Successive Squaring Algorithm for computing a^k mod m.

Quite straightforward.

End up with 18 14 18 14 18 14 18 18 7 10 7 7 10 14 7 10 14 18 10 as the non-zero remainders that are multiplied together to give 439205094557429760000 mod 31 which is 10.
Will do so demain in the morning, thanks.

17. Hey, was just wondering if I've done this question correctly for part b?

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@zacken
18. Also, surely Fermat's Little theorem would be easier to apply to work out part c?
@Zacken fishyfishy1999
19. (Original post by maths10101)
Also, surely Fermat's Little theorem would be easier to apply to work out part c?
@Zacken fishyfishy1999

FLT is often used in tandem with successive squaring to checking if a number is composite without finding the factors themselves.
20. @fishyfishy1999 it works perfectly here..get the same answer

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