Show that: Show that there is odd number of elements of a finite group satisfying $x^3=e $? And even number of elements satisfying $x^2\neq e$???
I donot have any idea how to start.
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The claim is false as can be seen by considering the group of order $3$. – Tobias Kildetoft May 03 '16 at 09:53
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1Do you mean "the number of elements such that $g^3=e$"? (that's different, as it includes the identity) – lulu May 03 '16 at 09:54
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What exactly do you mean by "order"? Most people when referring to "order of a group $G$ refer to the size/cardinality. of the set $|G|$. – Siddharth Bhat May 03 '16 at 09:57
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3Hint: if $x,y$ are two such elements ($\neq e$) then the two sets ${e,x,x^2}$ and ${e,y,y^2}$ either coincide or intersect only at $e$. – lulu May 03 '16 at 10:06
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I have edited the question/ – Rayees Ahmad May 03 '16 at 10:15
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There is an involution on the set of all elements $g$ such that $g^3=e$ sending $g$ to $g^{-1}$. This involution has exactly one fixed point, namely $g=e$. Hence altogether there is an odd number of them.
Mikhail Katz
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2Look at it this way: if you have an innumerable number of pairs of ears, and in addition a nose, then altogether you will necessarily have an odd number of organs. – Mikhail Katz May 03 '16 at 11:28
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1The nose is the neutral element (read "Perfume: The Story of a Murderer", of Patrick Süskind). – Piquito May 03 '16 at 11:54
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@MikhailKatz Is there anything in the proposition that requires finiteness of group ? – manifold Jun 24 '18 at 16:31
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1No, only the finiteness of the number of elements $g$ satisfying $g^3=e$. @viru – Mikhail Katz Jun 25 '18 at 08:58
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Let $M = \{x \in G-\{e\}, x^3=e\}$. It is easy to show that if $x \in M$ then $x^{-1} \in M$, because $x^{-1}=x^2$. Since $M=\bigcup \{x, x^{-1}\}$ where the sets $\{x, x^{-1}\}$ are disjoint, we deduce $M$ has an even number of elements. It is important to notice that if $x \in M$ then $x^{-1} \ne x$, otherwise $x=x^{-1}$ implies $x=x^2$, then $x=e$.
For the second part, let $P=\{ x \in G, x^2 \ne e \}$. If $x \in P$ then $x^{-1} \in P$, otherwise $x^2 = e$. Then $P=\bigcup \{x, x^{-1}\}$ where the sets $\{x, x^{-1}\}$ are disjoint, so $P$ has an even number of elements.