Let $G$ be the group of all the maps from closed interval $[0,1]$ to $\mathbb{Z}$. The subgroup $H= \left \{ f \in G :f(0)=0 \right \}$


$1)$ $H$ is countable

$2)$ $H$ is uncountable

$3)$ $H$ has countable index

$4)$ $H$ has uncountable index

Solution I tried- In this question he is asking about maps not for functions. The number of maps form $[0,1]$ to $\mathbb{Z}$ must be more than $\aleph_0^\mathfrak{c}$. Now I am confused here, how to proceed further because, I have no idea what is $\mathfrak{c}$ times $\aleph_0$. please give me a hint so that I can solve this further.

Thank you.

  • $\begingroup$ $\mathfrak c$ "times" $\aleph_0$ is $\mathfrak c$. $\endgroup$ – Rick Jun 15 '20 at 9:21
  • $\begingroup$ so the cardinality of $G$ is $c$ but about $H$ can you please help? $\endgroup$ – honey kumar Jun 15 '20 at 9:22
  • $\begingroup$ What is your definition of "map"? If it doesn't mean "function" this needs to be clarified. $\endgroup$ – diracdeltafunk Jun 15 '20 at 9:53

Consider the map $\varphi\colon G\to\mathbb{Z}$ defined by $\varphi(f)=f(0)$.

Then this map is a group homomorphism (verify it), it is surjective (just consider constant functions) and $H=\ker\varphi$.

Therefore by the homomorphism theorem, $G/H=G/\ker\varphi\cong\mathbb{Z}$.

This answers the question about the index, doesn't it?

Also, since you have just fixed the image at $0$, you can easily seen that $H$ is isomorphic to the group of all function $(0,1]\to\mathbb{Z}$, which has cardinality $\aleph_0^{\mathfrak{c}}=2^{\mathfrak{c}}$.


Since the set of all the binary sequences (i.e. functions with domain $\mathbb{N}$ and range in $\{ 0, 1 \}$) is uncountable, it can be shown that $H$ is also uncountable.


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