In (π, +) , let π» = π ππ‘ ππ πππ ππ’ππ‘πππππ ππ 3 and
πΎ = π ππ‘ ππ πππ ππ’ππ‘πππππ ππ 5.
Show that H and K are subgroups of Z . Also describe π» β© πΎ
We shall show that H and K are closed under + and every element in H and K have their respective inverses in H and K
Let "h,h\u00b0\u2208H" . Then "h=3k" and "h\u00b0=3q" , where "k,q\u2208\u2124"
"h+h\u00b0=3k+3q=3(k+q)"
let "k+q=m\u2208\u2124"
Hence, H is closed under +
Let "h\u00b0" be the inverse of "h\u2208H"
"=>h+h\u00b0=0" (identity element)
"=>3k+h\u00b0=0"
"=>h\u00b0=-3k=3(-k)"
Let "-k=n\u2208\u2124"
Thus, "h\u00b0\u2208H"
Hence, "(H,+)" is a subgroup of "(\u2124,+)"
Let "k,k\u00b0\u2208H" . Then "k=5r" and "k\u00b0=5q" , where "r,q\u2208\u2124"
"k+k\u00b0=5r+5q=5(r+q)"
let "r+q=m\u2208\u2124"
Hence, K is closed under +
Let "k\u00b0" be the inverse of "k\u2208H"
"=>k+k\u00b0=0" (identity element)
"=>5r+k\u00b0=0"
"=>k\u00b0=-5r=5(-r)"
Let "-r=n\u2208\u2124"
Thus, "k\u00b0\u2208H"
Hence, "(K,+)" is a subgroup of "(\u2124,+)"
Hβ©K={zββ€: z=15m, mββ€}
Comments
Leave a comment