Let $G$ be the group of all the maps from closed interval $[0,1]$ to $\mathbb{Z}$.

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

Then

$$1)$$ $$H$$ 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.

• $\mathfrak c$ "times" $\aleph_0$ is $\mathfrak c$. – Rick Jun 15 '20 at 9:21
• so the cardinality of $G$ is $c$ but about $H$ can you please help? – honey kumar Jun 15 '20 at 9:22
• What is your definition of "map"? If it doesn't mean "function" this needs to be clarified. – 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, $$G/H=G/\ker\varphi\cong\mathbb{Z}$$.
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.