# Show $\forall a \notin H$, $3\mid\mathrm{ord}(a)$, where $H$ is a normal subgroup of order $8$, $G$ is a group of order $24$

Let $G$ be a group with $24$ elements. Let $H$ be a normal subgroup of $G$ s.t. $|H|=8$. Show that $\forall a \notin H$, $3\mid\mathrm{ord}(a)$.

I know that order of $aH$ in quotient group $G/H$ has order $3$, that is $a^3 \in H$, but I just don't know if it is enough to say $3\mid\mathrm{ord}$. Since if $\mathrm{ord}(a^3)=n$ and $a^{3 n}=1$, maybe $a^n=1$.

Thank you.

• $\Bbb Z_{24}$ has $H = 3\Bbb Z_{24}$ as a normal subgroup of order $8$. $\text {order}(1) = 24$, which does not divide $3$. Perhaps you mean $3 | \text{order}(a)$? – Paul Sinclair Oct 8 '15 at 23:38
• @PaulSinclair Oh yea. Let me correct it. – k99731 Oct 8 '15 at 23:43
• I mostly avoid group theory questions because my group theory knowlege is so rusty, but I think you should consider the order of $ah$, for $h \in H$. What happens if $3$ doesn't divide the order of $a$ ? – Paul Sinclair Oct 8 '15 at 23:56
• @PaulSinclair Oh I think I got it. If order of $a =n$, $a^n=(a 1)^n=1$ must be in $H$. Then $a^n H=H$, then $3|n$? – k99731 Oct 9 '15 at 0:14
• I don't think it is immediate that $a^nH = H$ implies $3|n$ – Paul Sinclair Oct 9 '15 at 0:28

For any $a\notin H$, $a^3H=(aH)^3=H$ for $|G/H|=24/8=3$. So $a^3\in H$. Suppose order of $a^3$ is $n$. Then $n|8$. So $n$ is prime to $3$. Thus $(a^3)^n=a^{3n}=1$, i.e. the order of $a$ is $3n$.
The order of $a$ can not be less than $3n$ for $n$ is prime to $3$. It can not be $n$ since otherwise, $a^n=1$ and $a\in H$ for $n|8$.