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Be group order pq where pfMe number Prove that any non-trivial subgroup of G cyclic: (Hint: Use Lagrange Theorem and the corollaries given the handout:...

Question

Be group order pq where pfMe number Prove that any non-trivial subgroup of G cyclic: (Hint: Use Lagrange Theorem and the corollaries given the handout:

be group order pq where pfMe number Prove that any non-trivial subgroup of G cyclic: (Hint: Use Lagrange Theorem and the corollaries given the handout:



Answers

Prove that when A is an m × n matrix, and B, C are n × p matrices, [3]
A(B + C) = AB + AC .

Hello. Real question. Envisages when that F B and ordered field and X. So I said enough. Okay. It has also given that if X less than zero and why less than that then we need to prove that X. Y greater than access it. So let us get to hear that if access less than zero, this can be written as minus Act should be greater than zero. Okay, now here, if y is less than that so Zach minus Y should be greater than zero. Okay, no, these two have become positive quantities. Some multiplication of two positive quantities should be always positive, should always be positive. So we stretch it as minus X. Which is a positive quantity. Now into that minus Y. Which is again a positive wants to know should be positive. Let us open the bracket minus X. Z bless X. Y should be positive. Let us add except to both the sides will be having X way this is minus exceed all. It is minus except plus X. Y. And we are adding acceptable the sides greater than exit. So these two will become zero. So from here we are getting X. Y greater than X zet. So this is the thing we need to prove. Thank you.

Hello. Real question. Envisages when that F B and ordered field and X. So I said enough. Okay. It has also given that if X less than zero and why less than that then we need to prove that X. Y greater than access it. So let us get to hear that if access less than zero, this can be written as minus Act should be greater than zero. Okay, now here, if y is less than that so Zach minus Y should be greater than zero. Okay, no, these two have become positive quantities. Some multiplication of two positive quantities should be always positive, should always be positive. So we stretch it as minus X. Which is a positive quantity. Now into that minus Y. Which is again a positive wants to know should be positive. Let us open the bracket minus X. Z bless X. Y should be positive. Let us add except to both the sides will be having X way this is minus exceed all. It is minus except plus X. Y. And we are adding acceptable the sides greater than exit. So these two will become zero. So from here we are getting X. Y greater than X zet. So this is the thing we need to prove. Thank you.

Reflexive and transitive than ours? Asymmetric. I think this is the most straightforward of the three Bruce. Okay, so uh let R. B. Your reflexive and transitive. Um oh let X Y bien are and now what I'm going to do is I'm going to suppose that Y X I suppose why X is in our then since our is transitive, X. X is in our and why why is in our but that's a contradiction are is mhm. It reflexive. So Y X. What I supposed right here cannot be in our and thus our is anti symmetric. Oh, no, no, no. Asymmetric, sorry. Yeah.

And it is given to us that and uh order of a. Yes I mean to him and or no BNC. It's and in two. Okay we know that but tradition. Mhm. We'll see what oh same order only. So after leading the N. C. We get the orders that is and into No we have to multiply with the the so you can see that this product is possible. Yes. No number Oh all of in a is it close to No not at all go in the floor here. Number of problem in A and and I'm gonna going to be placido and the product is possible. And after making the product can see that in order that was a family too well and then into a after multiplying them we get order of them into okay so we can say that kind of no order at all angel, let's see. Mhm. And growth. No we should move to the right side. In the right side. There is Eddie lot it's we can see that order eight I'm into an order of these and into the similarly order of age. I mean well order of these and into after making the product we can see that we remain with I'm going to be and we have to add yeah the product O. N. C. And it will be also I mean to and all that and due to this event on order after adding them the guilt again. I'm sorry I'm going to be. I mean so we can see that the orders in both the case here also it is I'm going to be and he had also did them into so we can say came to be. Let's see it, baby. Plus it's okay. Thank you.


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