Let f:(Z28,+)→(Z16,+) be a group homomorphism such that f(1)=12 Find ker(f)

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Let  f:(Z28,+)→(Z16,+)  be a group homomorphism such that f(1)=12  Find ker(f)

John Koga · Accepted Answer

Step 1  f:(Z28,+)→(Z16,+) be a group homomorphism.  Since, 1 is a generator of Z28,Im(f)=⟨f(1)⟩=⟨12⟩  cyclic subgroup of Z16  |Im(f)|=ord(12)=4  Then by first isomorphism theorem,  Z28ker(f)=∼Im(f)  Hence, |ker(f)|=284=7  So, ker(f) is a cyclic subgroup of Z28 generated by an element of order 7.  And, Z28 and element of order 7 is 4.  Hence, ker(f)=⟨4⟩

encolatgehu · Accepted Answer

Step 1  We have  12+12(mod 16)=f(1)+f(1)  =f(1+1)  =8(mod 16)  8+12 (mod 16)=f(2)+f(1)  =f(2+1)  =f(3)  =4 (mod 16)  and 4+12 (mod 16)=f(3)+f(1)  =f(3+1)  =f(4)  =0 (mod 16)  Thus kerf=⟨4⟩

alenahelenash · Accepted Answer

Step 1 Observe that f(2)=8≠0, f(7)=4≠0, f(1)=12≠0 and f(4)=0 so the answer is ker⁡f=&amp;lt;&amp;lt;4&amp;gt;&amp;gt;.

Let f:(\mathbb{Z}_{28},+)\rightarrow(\mathbb{Z}_{16},+) be a group homomorphism such that f(1)=12 Find \ker(f)

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