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

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VincentCheung
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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!Image
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0x2a
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Are you sure  \mathbb{Z}^n is not the direct product of  \mathbb{Z} n times?

If  \mathbb{Z}^n = \mathbb{Z}/n\mathbb{Z} then the problem doesn't even make sense, as there are  2^n functions (NOT homomorphisms) from \mathbb{Z}/n\mathbb{Z} to \mathbb{Z}_2, and by taking away the constant functions f: \mathbb{Z}/n\mathbb{Z} \rightarrow \{0\} and f:\mathbb{Z}/n\mathbb{Z} \rightarrow \{1\} you are left with 2^n - 2 surjective functions, which is far greater than the number of possible homomorphisms.
In fact, a homomorphism from  \mathbb{Z}/n\mathbb{Z} when n is odd to \mathbb{Z}_2 doesn't even exist.

For the second part, from the first isomorphism theorem you know that  G/Ker \varphi \cong Im\varphi, so any homormorphism that sends G to \{0,1\} must have an index of two. Also show the converse, that any group of index two is a kernel of some isomorphism from \mathbb{Z}^n \rightarrow \mathbb{Z}_2.
Hint:
Spoiler:
Show
Subgroups of index 2 are normal
.
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VincentCheung
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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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(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  \mathbb{Z}^n as a vector space (well in this case, more of a free module), the standard basis for \mathbb{R}^n also spans  \mathbb{Z}^n and is linearly independent. Now you just need to send each  e_i = (0,...,1,...,0) with 1 in the ith place to a 0 or 1 as  \mathbb{Z}^n 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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VincentCheung
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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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(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 e_is can map to 0. Otherwise \mathbb{Z}^n would map to just \{0\} because e_is generate \mathbb{Z}^n and our mapping is a homomorphism, so our map would no longer be surjective.
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VincentCheung
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Makes perfect sense now, thanks very much for your help!
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