# Thread: isomorphisms between certain groups

1. ## isomorphisms between certain groups

It is possible to show easily that $\displaystyle \mathbb{Z} / 2$ is isomorphic with the group of units denoted $\displaystyle U_3$ that have a multiplicative inverse modulo 3 .
defined by the isomorphic map : $\displaystyle \varphi : \mathbb{Z}/2 \rightarrow \ U_3$
$\displaystyle \varphi (0) = 1$, $\displaystyle \varphi (1) = 2$

Furthermore we can find another isomorphic map to show that $\displaystyle \phi : U_5 \rightarrow \mathbb{Z}/2 \times \mathbb{Z}/2$

I cannot really prove that the map $\displaystyle \phi : U_5 \rightarrow \mathbb{Z}/4$ cannot be an isomorphism. However i can define the map $\displaystyle \phi (m) = m-1$ and by counter-example to prove that this is not isomorphic as
$\displaystyle \phi(4)=\phi(2) + \phi(2)$ if it is an isomorphism
but $\displaystyle = 1 + 1 = 2$ which does equal 3

However this is not very rigorous and I would like to know if there a deeper/more rigorous way of proving that there cannot be an isomorphism

2. An easy way is to look at the orders of the elements, keeping in mind that isomorphisms preserve orders. In particular, the order of every element in $\displaystyle U_5$ is at most two, whereas some elements in $\displaystyle \mathbb{Z}_4$ have order four. Therefore, the two groups cannot be isomorphic.

3. Originally Posted by FGT12
It is possible to show easily that $\displaystyle \mathbb{Z} / 2$ is isomorphic with the group of units denoted $\displaystyle U_3$ that have a multiplicative inverse modulo 3 .
defined by the isomorphic map : $\displaystyle \varphi : \mathbb{Z}/2 \rightarrow \ U_3$
$\displaystyle \varphi (0) = 1$, $\displaystyle \varphi (1) = 2$

Furthermore we can find another isomorphic map to show that $\displaystyle \phi : U_5 \rightarrow \mathbb{Z}/2 \times \mathbb{Z}/2$

I cannot really prove that the map $\displaystyle \phi : U_5 \rightarrow \mathbb{Z}/4$ cannot be an isomorphism. However i can define the map $\displaystyle \phi (m) = m-1$ and by counter-example to prove that this is not isomorphic as
$\displaystyle \phi(4)=\phi(2) + \phi(2)$ if it is an isomorphism
but $\displaystyle = 1 + 1 = 2$ which does equal 3

However this is not very rigorous and I would like to know if there a deeper/more rigorous way of proving that there cannot be an isomorphism
To add on to roninpro's explanation it's a common fact that every group of order $\displaystyle p^2$ is abelian and thus by the structure theorem isomorphic to either $\displaystyle C_p\oplus C_p$ or $\displaystyle C_{p^2}$. So, just noting that $\displaystyle \left|U_5\right|=\phi\left(5\right)=5-1=4$ narrows it down to two groups up to isomorphism. You can then extend what roninpro's one-hundred percent correct explanation to show (and I'm urging you to do this!) that $\displaystyle C_{mn}\cong C_m\oplus C_n$ if and only if $\displaystyle (m,n)=1$!