On a group
- Tolaso J Kos
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On a group
Let $(G, *)$ be a group. If $|G|=2n$ , then prove that there exists an element $a \neq e_G$ in $G$ such that $a*a=e_G$ , where $e_G$ is the identical element of $G$.
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- Grigorios Kostakos
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Re: On a group
Because $2$ is a prime factor of $2n=|G|$, by Cauchy's theorem, we have that $G$ has an element of order $2$, i.e. exists $a\in G\setminus\{e_G\}$ such that $a\ast a= e_{G}$ .
Grigorios Kostakos
Re: On a group
This exercise is trivial using Cauchy's theorem, which is kind of an overkill in this situation. Here is a more elementary proof:
The idea is to pair every element of the group with its inverse. Setting aside the identity element, $2n-1$ elements remain, which is an odd number of elements. So we seperate them as follows: there are at most $n-1$ distinct elements, along with their $n-1$ inverses and at least $1$ more, which is equal to its own inverse (it cannot be one of the other elements, an inverse of an element, or equal to $1_G$, by choice). Calling $a$ an element with this property, we get: $a=a^{-1}\Leftrightarrow a^2=1_G$.
For better understanding, try writing down the elements of the group. For example, if there is only one such element $a$ (element of order 2) we can write: $G=\left \{1,a,a_{1},a_{1}^{-1},...,a_{n-1},a_{n-1}^{-1} \right \}$, where $a_{i}$ are the rest of the elements. Of course $a$ doesn't need to be unique.
The idea is to pair every element of the group with its inverse. Setting aside the identity element, $2n-1$ elements remain, which is an odd number of elements. So we seperate them as follows: there are at most $n-1$ distinct elements, along with their $n-1$ inverses and at least $1$ more, which is equal to its own inverse (it cannot be one of the other elements, an inverse of an element, or equal to $1_G$, by choice). Calling $a$ an element with this property, we get: $a=a^{-1}\Leftrightarrow a^2=1_G$.
For better understanding, try writing down the elements of the group. For example, if there is only one such element $a$ (element of order 2) we can write: $G=\left \{1,a,a_{1},a_{1}^{-1},...,a_{n-1},a_{n-1}^{-1} \right \}$, where $a_{i}$ are the rest of the elements. Of course $a$ doesn't need to be unique.
Re: On a group
Another solution:
We define $X_1=\left \{ 1_G \right \}, X_2=\left \{ x\in G: x\neq 1_G, x=x^{-1} \right \}$ and $X_3=G\setminus (X_1\cup X_2)$ is the set of the rest of the group's elements. These sets make up a partition of $G$ due to the uniqueness of element order and thus $G$ can be written as their disjoint union, namely $G=X_1\cup X_2\cup X_3$. Summing, we pass to orders and get $2n=1+\left | X_2 \right | + \left | X_3 \right |$, so the sum $\left | X_2 \right | + \left | X_3 \right |$ is odd, which means that one of the terms is odd and the other is even. In particular, $\left | X_3 \right |$ is even, since it can be further partitioned by sets of the form $\left \{ a,a^{-1} \right \}$, where $a\neq a^{-1}$ in $X_3$. Since $G$ has even order, we conclude that $X_2$ has odd order. This means that $X_2\neq \varnothing$ and this ensures the existence of such an element.
We define $X_1=\left \{ 1_G \right \}, X_2=\left \{ x\in G: x\neq 1_G, x=x^{-1} \right \}$ and $X_3=G\setminus (X_1\cup X_2)$ is the set of the rest of the group's elements. These sets make up a partition of $G$ due to the uniqueness of element order and thus $G$ can be written as their disjoint union, namely $G=X_1\cup X_2\cup X_3$. Summing, we pass to orders and get $2n=1+\left | X_2 \right | + \left | X_3 \right |$, so the sum $\left | X_2 \right | + \left | X_3 \right |$ is odd, which means that one of the terms is odd and the other is even. In particular, $\left | X_3 \right |$ is even, since it can be further partitioned by sets of the form $\left \{ a,a^{-1} \right \}$, where $a\neq a^{-1}$ in $X_3$. Since $G$ has even order, we conclude that $X_2$ has odd order. This means that $X_2\neq \varnothing$ and this ensures the existence of such an element.
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