Finding the kernel of $\phi$ of applying the First Isomorphism Theorem Let $\mathbb T$={$z\in$ $\mathbb Z$ : $\vert z \vert$=$1$}Define $\phi$:$\mathbb R$ $\to$ $\mathbb T$ by $\phi$($\theta$)=$e^{i\theta \pi}$, $\theta$ $\in$ $\mathbb R$. Find the kernel of $\phi$ and apply the FIT(First Isomorphism Theorem)  Definition of a kernel of $\phi$: ker $\phi$ :={$x\in G_1$ : $\phi (x)$= $e_2$} note: $\phi^{-1}$($e_2$) is also part of the defintion. $G_1$ is a group and $e_2$ is an identity in the group Definition of First Isomorphism Theorem:  Let $G_1$ $\to$ $G_2$ be groups and let $\phi$ : $G_1 \to G_2$ be a homomorphism. Then $\frac{G_1}{ker \phi}$ is homomorphic to $\phi$($G_1$)Definition of homomorphism: Let $G_1$ and $G_2$ be two groups. Then $\phi$ : $G_1 \to G_2$ is called a homomorphism iff $\forall$ $a,b \in G_1$. $\phi$($ab$)= $\phi$($a$) $\phi$($b$) note: $\mathbb T$ is a subgroup of the multiplicative group of non-zero complex numbers. I need help on the First Isomorphism Theorem and probably the ker $\phi$  Here is my attempt on the ker $\phi$ part and defining the function:$\phi$ is a homomorphism because $\phi$($x+y$)=$\phi$($x$)$\phi$($y$) by the addition formulas for $sin$ and $cos$.The ker $\phi$ consists of all $x\in$ $\mathbb R$ so $e^{i\theta \pi}$=$cos\pi \theta$ + $isin\pi \theta$=$1$then $cos\pi \theta$=$1$ and $sin\pi \theta$=$0$but $cos\pi \theta$=$1$ and it forces $x$ to be an integer since $1\in$ ker $\phi$ We have ker $\phi$=$\mathbb Z$By the FITsince ker $\phi$=$\mathbb Z$Then $\frac{\mathbb R}{\mathbb Z}$ is a subgroup of $\mathbb T$ Did I do this right?
 A: Alright, The first isomorphism theorem (FIT) states that if we have an action $\phi: G\to H$ that is homomorphic, then the $ker(\phi)$ is a normal subgroup of $H$. 

First, we need to show that $\phi:\mathbb R\to \mathbb T$ such that $\phi (\theta)=e^{i\pi\theta}$ where $\mathbb T:= \{z\in \mathbb C:\;||z||=1\}$. This can be easily done by letting $x,y\in \mathbb R$, So $\phi(x+y)=e^{i\pi(x+y)}=(e^{i\pi x})e^{i\pi y}=\phi(x)\phi(y)$.
Next, we need to find the kernal of $\phi$. This can be done by noting that $1$ is the identity of $\mathbb T$ so let $$\phi(\theta)=e^{i\pi\theta}=\cos(\pi\theta)+i\sin(\pi\theta)=1+0i.$$
This implies that $\cos(\pi\theta)=1$. As such $\theta = 2k$, $\forall k \in \mathbb Z$. So, 
$$ker(\phi)=\{n=2k: k \in \mathbb Z\}$$
Lastly, we need to show that $ker(\phi)$ is a normal subgroup of $\mathbb T$ this is done by showing that $gH=Hg$ if $g\in ker(\phi)$. Thus,
$$\phi(n)H=e^{i\pi n}e^{i\pi\theta}=e^{i(2\pi) k}e^{i\pi\theta}=e^{i\pi\theta}=e^{i\pi\theta}e^{i(2\pi) k}=H\phi(n) $$
as desired.
A: The kernel of a morphism is the set of elements which map to the identity in the range. Here, the identity is 1. So you're trying to find $\theta$ such that $e^{i\theta} = 1$. Well $e^{i\theta} = cos(\theta) + isin(\theta)$ so we're trying to find $\theta$ such that $cos(\theta) + isin(\theta) = 1$.
You're close up above in the comments, but $cos(\pi) = -1$. The absolute value (modulus) isn't used here, it's just used to define the elements of $\mathbb{T}$.
Also the first isomorphism theorem for groups says that given a homomorphism $\phi : A \to B$, $A / ker(\phi) \cong B$, not that $A / ker(\phi)$ is a subgroup of $B$ (well it is in a sense, as it's the whole thing). To understand more what's happening, look up quotient groups.
