# Element of Center in Group whose Order is Power of 2

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## Theorem

Let $n \in \Z$ be an integer such that $n \ge 2$.

Let $G$ be a group whose order is $2^n$.

Let $x \in G$ be of order $2^{n - 1}$ in $G$.

Then $x^{2^{n - 2} }$ is an element of the center of $G$.

## Proof

Let $H = \gen x$ be the subgroup of $G$ generated by $x$.

We have that the index of $H$ in $G$ is $2$.

Then by Subgroup of Index 2 is Normal, $H$ is normal in $G$.

Let $y = x^{2^{n - 2} }$.

We have that $y \in H$, by definition of the construction of $H$.

Then the order of $y$ in $G$ is $2$.

As $H$ is normal, we have:

- $\forall g \in G: g y g^{-1} \in H$

By Order of Conjugate Element equals Order of Element, $g y g^{-1}$ is also of order $2$.

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $11$