# Thread: please check if my proof is correct

1. ## please check if my proof is correct

Check if $\langle t^{n} \rangle$ is a retract of $\langle t \rangle$.

Proof:
Define $\phi : \langle t \rangle \rightarrow \langle t \rangle$ by $\phi (x)=x$ for all $x \in \langle t \rangle$.

Let $x_{1}, x_{2} \in \langle t \rangle$.
Then $x_{1}=t^{a}, x_{2}=t^{b}$ where $a,b$ are integers.

$\phi (x_{1}x_{2})= \phi (t^{a}t^{b})=\phi (t^{a+b})=t^{a+b}=t^{a}t^{b}=\phi (t^{a} \phi (t^{b})=\phi (x_{1}) \phi(x_{2})$
So, $\phi$ is homomorphism.

Clearly, $\phi (x^*)=x^*$ for all $x^* \in \langle t^{n} \rangle$.

Since $\phi (x_{1})= \phi (t^{a})=t^{a}$, $t^{a} \in \langle t^{n} \rangle \Leftrightarrow a | n$.

Hence, $\langle t^{n} \rangle$ need not be a retract of $\langle t \rangle$.

2. Originally Posted by deniselim17
Check if $\langle t^{n} \rangle$ is a retract of $\langle t \rangle$.

Proof:
Define $\phi : \langle t \rangle \rightarrow \langle t \rangle$ by $\phi (x)=x$ for all $x \in \langle t \rangle$.

Let $x_{1}, x_{2} \in \langle t \rangle$.
Then $x_{1}=t^{a}, x_{2}=t^{b}$ where $a,b$ are integers.

$\phi (x_{1}x_{2})= \phi (t^{a}t^{b})=\phi (t^{a+b})=t^{a+b}=t^{a}t^{b}=\phi (t^{a} \phi (t^{b})=\phi (x_{1}) \phi(x_{2})$
So, $\phi$ is homomorphism.

Clearly, $\phi (x^*)=x^*$ for all $x^* \in \langle t^{n} \rangle$.

Since $\phi (x_{1})= \phi (t^{a})=t^{a}$, $t^{a} \in \langle t^{n} \rangle \Leftrightarrow a | n$.

Hence, $\langle t^{n} \rangle$ need not be a retract of $\langle t \rangle$.

What is t, anyway? A normal subgroup of a group is a retract of the group iff it has a normal complement.

Tonio

3. $\langle t \rangle$ is a cyclic group generated by $t$.
$t$ is a generator.

4. Originally Posted by deniselim17
$\langle t \rangle$ is a cyclic group generated by $t$.
$t$ is a generator.

Good. So <t> is a cyclic group and thus it cannot be expressed as a non-trivial direct product ==> no subgroup of <t> has normal complement ==> there exists no retract of <t>.

Tonio