# Show there are 2n − 1 surjective homomorphisms from Zn to Z2, 1st Isomorphism thm

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#1
I am having trouble with the following problem. I have tried to use induction for the first part however I am unsure if I have used the induction hypothesis correctly.

I have said that up to k-1 we assume it is true and then with the next n = k i have said that this only introduces a new term which can be mapped to either 0 or 1, then it's just 2.2k -1 -1 which is and we're done. However I'm not entirely sure about my reasoning, also i don't see how the first isomorphism theorem..

Any help is much appreciated!
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5 years ago
#2
Are you sure is not the direct product of n times?

If then the problem doesn't even make sense, as there are functions (NOT homomorphisms) from to , and by taking away the constant functions and you are left with surjective functions, which is far greater than the number of possible homomorphisms.
In fact, a homomorphism from when n is odd to doesn't even exist.

For the second part, from the first isomorphism theorem you know that , so any homormorphism that sends to must have an index of two. Also show the converse, that any group of index two is a kernel of some isomorphism from .
Hint:
Spoiler:
Show
Subgroups of index 2 are normal
.
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#3
my bad, read the question wrong. Using the isomorphism theorem now makes sense, thanks.

However the first part I am still confused about which method to use to show it
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5 years ago
#4
(Original post by VincentCheung)
my bad, read the question wrong. Using the isomorphism theorem now makes sense, thanks.

However the first part I am still confused about which method to use to show it
For the first part just think of as a vector space (well in this case, more of a free module), the standard basis for also spans and is linearly independent. Now you just need to send each with 1 in the ith place to a 0 or 1 as is an additive group generated by the standard basis.

Of course, the above reasoning shouldn't be the solution, because introducing some more complex algebraic structure to solve a problem about groups is cheating.
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#5
Ah okay, I can see where the 2^n comes from now, just struggling to see why the -1 comes in.. I've tried to do it for cases n=1,2,3 but still about setting up the maps so that this is true.
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5 years ago
#6
(Original post by VincentCheung)
Ah okay, I can see where the 2^n comes from now, just struggling to see why the -1 comes in.. I've tried to do it for cases n=1,2,3 but still about setting up the maps so that this is true.
-1 comes from the fact that not all of the s can map to 0. Otherwise would map to just because s generate and our mapping is a homomorphism, so our map would no longer be surjective.
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#7
Makes perfect sense now, thanks very much for your help!
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