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Groups isomorphisms watch

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1. Hey just real confused on parts a,b and c??
Any ideas?
Any response/feedback will be very much appreciated thanks

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2. @Zacken any ideas mate?

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3. (Original post by maths10101)
Hey just real confused on parts a,b and c??
Any ideas?
Any response/feedback will be very much appreciated thanks

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Part a) is using the definition of a group isomorphism, the fact if, a is the identity in G, then f(a) is the identity in H. Which should lead into part b and c.

(Sorry if not very helpful or if it is the wrong way to go at it )
4. (Original post by zetamcfc)
Part a) is using the definition of a group isomorphism, the fact if, a is the identity in G, then f(a) is the identity in H. Which should lead into part b and c.

(Sorry if not very helpful or if it is the wrong way to go at it )
Hmm, so if a is the identity in G, then it has order 1. Thus f(a) will give the identity in H, and then the order of f(a) is simply 1 again, so this meaning o(a)=o(f(ah))??
Is that what you're getting at?

...And then you would just prove it for arbitrary a and b of H, such that o(a+b)=o(f(a)+f(b))? Etc, basically applying the order to the ring axioms with order being n?...

^not sure if that top part makes sense lol

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5. (Original post by maths10101)
Hmm, so if a is the identity in G, then it has order 1. Thus f(a) will give the identity in H, and then the order of f(a) is simply 1 again, so this meaning o(a)=o(f(ah))??
Is that what you're getting at?

...And then you would just prove it for arbitrary a and b of H, such that o(a+b)=o(f(a)+f(b))? Etc, basically applying the order to the ring axioms with order being n?...

^not sure if that top part makes sense lol

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Well if an arbitrary number of binary operations give you the identity in G then as f(a)#f(b) is the same as f(a#b) this number of binary operations in H will give you the identity in H. Therefore the orders of the element in G is the same as its image in H.v(using # as a binary operation)

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