Hw8 Problem 11

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Problem 11

(*) Let $|G| = 8$. Show that $G$ must have an element of order $2$.

Solution

Let $|G| = 8$. Subgroups of $G$ must have order that divides $8$, so then $1, 2, 4,$ or $8$. Only the identity element has order $1$, so there must be at least one different order. If $G$ has an element $g$ of order $4$, then $g ^ {4} = e$, so $(g ^ {2}) ^ {2} = e$, so $g ^ {2}$ has order $2$. If $G$ has an element $h$ of order $8$, then $h ^ {8} = e$, or $(h ^ {4}) ^ {2} = e$, so $h ^ {4}$ has order $2$. If $G$ does not contain an element of order $4$ or $8$, then all non-identity elements have order $2$. Therefore if $|G| = 8$, then $G$ must have an element of order $2$.