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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 FIT
since ker $\phi$=$\mathbb Z$
Then $\frac{\mathbb R}{\mathbb Z}$ is a subgroup of $\mathbb T$

Did I do this right?

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  • $\begingroup$ $\cos \pi = -1$. $\endgroup$ – Qiaochu Yuan Mar 17 '17 at 3:31
  • $\begingroup$ I used $\theta$ with it too $\endgroup$ – behold Mar 17 '17 at 3:32
  • $\begingroup$ You say that the kernel is $\mathbb{Z}$, but that would imply that $e^{i \pi} = \cos \pi + i \sin \pi = 1$, which is false, because $\cos \pi = -1$. $\endgroup$ – Qiaochu Yuan Mar 17 '17 at 3:33
  • $\begingroup$ you leaving out theta $\endgroup$ – behold Mar 17 '17 at 3:35
  • $\begingroup$ I was using the absolute value from my first statement to get a positive number $\endgroup$ – behold Mar 17 '17 at 3:37
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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.

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    $\begingroup$ Oh forgot the fact that using the (FIT) we have $\phi: G/N\to H$ is isomorphic where $N$ is the normal subgroup of $G$. So $\mathbb R /ker(\phi) \cong \mathbb T$. $\endgroup$ – Sentinel135 Mar 18 '17 at 1:48
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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.

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  • $\begingroup$ So if I change it to $-1$ I can still show it? I got show that kernel is equal to the set of $\mathbb Z$ to complete it $\endgroup$ – behold Mar 17 '17 at 4:23
  • $\begingroup$ Whoops I misread the function, we want $e^{i\pi\theta} = 1$, so you want $\theta$ such that $cos(\pi\theta) + isin(\pi\theta) = 1$. The answer still isn't $\mathbb{Z}$ again if $\theta = 1$, then $e^{i\pi} = -1$. $\endgroup$ – Kit Mar 17 '17 at 4:24
  • $\begingroup$ I have quotient groups in front of me. I could not tag it because of the limits but the quotient group deals with the left cosets $\endgroup$ – behold Mar 17 '17 at 4:26
  • $\begingroup$ I see @kit. This problem is difficult to me $\endgroup$ – behold Mar 17 '17 at 4:27
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    $\begingroup$ You're very close! Check my edits to my comment above I misread the function. You just need to consider which elements will make $cos(\theta\pi) + isin(\theta\pi) = 1$. It's definitely a subgroup of $\mathbb{Z}$. If you want a second reference on group theory, check out Aluffi, it's a great introduction. $\endgroup$ – Kit Mar 17 '17 at 4:29

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