Answer to Question #313206 in Abstract Algebra for Jyo

Question #313206

In (𝑍, +) , let 𝐻 = 𝑠𝑒𝑑 π‘œπ‘“ π‘Žπ‘™π‘™ π‘šπ‘’π‘™π‘‘π‘–π‘π‘™π‘’π‘  π‘œπ‘“ 3 and


𝐾 = 𝑠𝑒𝑑 π‘œπ‘“ π‘Žπ‘™π‘™ π‘šπ‘’π‘™π‘‘π‘–π‘π‘™π‘’π‘  π‘œπ‘“ 5.


Show that H and K are subgroups of Z . Also describe 𝐻 ∩ 𝐾

1
Expert's answer
2022-03-18T01:12:59-0400

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,+)"



H∩K={zβˆˆβ„€: z=15m, mβˆˆβ„€}








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