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Sol3 - Algebra 3(2004-05 – Solutions to Assignment 3...

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Unformatted text preview: Algebra 3 (2004-05) – Solutions to Assignment 3 Instructor: Dr. Eyal Goren 1) If G,H are finite groups such that ( | G | , | H | ) = 1 then every group homomorphism f : G-→ H is trivial ( f ( G ) = { 1 } ). Proof. Let K be the kernel of f . Then | Im( f ) | = | G | / | K | divides both | H | and | G | . Hence | Im( f ) | = 1 and f is trivial. / 2) Find all possible homomorphisms Q-→ S 3 . Is there an injective homomorphism Q-→ S 4 ? (As usual, Q is the quaternion group of order 8). Solution: The image of such a homomorphism divides both | S 3 | = 6 and | Q | = 8. Thus, the image is either of cardinality 1 and then f is the trivial homomorphism, or of cardinality 2. In the latter case, the kernel of f is either h i i , h j i or h k i . If the homomorphism is not trivial then all the elements outside the kernel must go to the same element of order 2 in S 3 , which is a transposition. There are 3 such elements. Thus, we find that there are, besides the trivial homomorphism, 9 more homomorphisms; eachelements....
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Sol3 - Algebra 3(2004-05 – Solutions to Assignment 3...

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