| Posted on Friday, 24 October, 2003 - 08:06 pm: |
Can you help me answer this question. Suppose H and K are subgroups of a group G. i) Prove that HÇK is a subgroup. ii) Prove that if HÈK is a subgroup then H Í K or K Í H.
| Posted on Friday, 24 October, 2003 - 09:45 pm: |
Go to the definition of Group.
A group G is a set of elements together with a binary operation which together satisfy the four fundamental properties of closure, associativity, the identity property, and the inverse property.
Elements A, B, C, ... with binary operation between A and B denoted AB form a group if
1. Closure: If A and B are two elements in G, then the product AB is also in G.
2. Associativity: The defined multiplication is associative, i.e., for all A, B, C in G (AB)C = A(BC)
3. Identity: There is an identity element I (a.k.a. 1, E, or e) such that IA = AI = A for every element A in G.
4. Inverse: There must be an inverse or reciprocal of each element. Therefore, the set must contain an element B = A-1 such that AA-1 = A-1 A = I for each element of G.
Let's start with i) and prove the four properties hold for H n K -- In general, this is the way to prove a subset of a group is a subgroup:
Closure: Let A and B be elements of H n K. AB is an element of H because H is a group. AB is an element of K because K is a group. Since AB is an element of H and AB is an element of K, AB is an element of H n K.
Associativity: Let A, B, and C be in H n K. A, B, and C are elements of H, so (AB)C = A(BC).
Identity: I is an element of H and I is an element of K, so I is an element of H n K.
Inverse: Let A be an element of H n K. Since A is an element of H, A-1 (such that AA-1 = A-1 A = I) is an element of H. Similarly, A-1 is an element of K. So A-1 is an element of H n K.
Now, for ii), let's assume the negation of the statement to be proved, and try for a contradiction. Suppose H u K is a subgroup, and an element A is in H but not K, and an element B is in K but not H.
For future reference, note that A-1 is in H because A is in H, and B-1 is in K because B is in K.
AB is an element of H u K, so AB is an element of H or AB is an element of K.
If AB is an element of H, then A-1 AB is an element of H, so B is in H, a contradiction.
If AB is an element of K, then ABB-1 is an element of K, so A is in K, a contradiction.
| Posted on Saturday, 25 October, 2003 - 12:48 pm: |
Hi Graeme. Thanks very much for your help and time. For part ii) Once we have obtained the contradiction what can we conclude?
| Posted on Saturday, 25 October, 2003 - 01:01 pm: |
Finally, I am stuck in proving the following.
Prove that the following are abelian groups. What are their orders and which are cyclic?
i){z in
:z12 =1}
under multiplication.ii){z in
:z12 =1}
under multiplication.
| Posted on Saturday, 25 October, 2003 - 02:05 pm: |
(i) List the members of z12 =1 in each case
(ii) By reference to Graham?s group definitions above, prove that they do or do not form groups
(iii) If they do form a group, then are they abelian? That shouldn?t be too difficult to decide given that multiplication is commutative.
(iv) The orders can be calculated rapidly (see the definition here )
(v) Prove the groups are cyclic by identifying the group generator.
Andre
| Posted on Sunday, 26 October, 2003 - 01:13 am: |
James,
In answer to the question you asked me, if you assume the opposite of the statement to be proved -- that is, you assume a statement that is true if and only if the statement to be proved is false -- then if you derive a contradiction, you know that the original statement is true. This is a time-honored method of proving things, which, in this case is very easy.
In this case I split the proof into two cases -- the case when AB is an element of H, and the case when AB is an element of K, and derived a contradiction in each case. So the statement I assumed in the very beginning is false, so its opposite -- the thing we're trying to prove -- must be true.
Your questions and my answers (and Andre's answers) illustrate a basic approach to proving facts about things (like groups) that have strict definitions that rely on various properties. The approach is to verify each of the properties (in my first answer) or to assume the opposite of what you intend to prove, and then use the properties to derive a contradiction (in my second answer).
I hope this helps.