SOLUTION: for sets H and K, we define the inersection H intersect k by
H intersect k = {x|x in H and x in k}
show that if H <= G and k <= G, then H intersect K <= g. ( remember: <= den
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Question 44451This question is from textbook
: for sets H and K, we define the inersection H intersect k by
H intersect k = {x|x in H and x in k}
show that if H <= G and k <= G, then H intersect K <= g. ( remember: <= denotes "is a sub group of," not "is a subgroup of.")
This question is from textbook
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
or sets H and K, we define the inersection H intersect k by
H intersect k = {x|x in H and x in k}
show that if H <= G and k <= G, then H intersect K <= g. ( remember: <= denotes "is a sub group of," not "is a subgroup of.")
CASE 1....H=G AND K=G....
THEN IT IMPLIES THAT IF H1 IS ANY ELEMENT OF H THEN IT IS ALSO AN ELEMENT OF G AND HENCE K.
SO H1 WILL BE AN ELEMENT OF H INTERSECTION K
SO H I.S. K IS A SUBSET OF G.................I
SIMILARLY ,WE CAN PROVE THAT IF G1 IS ANY ELEMENT OF G THEN IT IMPLIES IT IS ALSO AN ELEMENT OF H ,K AND HENCE H I.S. K
SO G IS A SUB SET OF H I.S. K.............II
FROM I AND II H I.S. K=G
CASE 2 .....H < G......AND K < G
HENCE IF H1 IS AN ELEMENT OF H IT IMPLIES IT IS AN ELEMENT OF G.
BUT IT MAY OR MAY NOT BE AN ELEMENT OF K
IF IT IS THEN H1 IS AN ELEMENT OF H I.S.K...HENCE H I.S.K IS A SUB SET OF G
IF IT IS NOT THEN IT IS NOT AN ELEMENT OF H I.S.K..SO OK.
NOW THE OTHER WAY ROUND..
SINCE H
THERE WILL BE AT LEAST ONE ELEMENT OF G SAY G1 WHICH IS NOT AN ELEMENT OF H.
HENCE G1 WILL NOT BE AN ELEMENT OF H I.S. K
HENCE H I.S.K IS A SUBSET OF G
SIMILARLY WE CAN ARGUE FOR K....TO SHOW H I.S.K IS A SUBSET OF G
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