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Here is the steps for the order of an element $a$ is $n$:
1) $a^n=e$
2)$\forall$m$\in$$\mathbb{z}$, ($a^m=e\Rightarrow n\mid m$)
My question states that I follow these 2 steps to show $a$ has order $2n$ and $a^2$ must have order $n$
I am not sure how to prove this.

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    Can you double check to make sure you're accurately stating the problem given? – 2017-01-31
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    This is the problem. the 2 given steps and use them steps to prove $2n$ and $a^2$ – 2017-01-31
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    "Prove $2n$ and $a^2$"? Those aren't statements to be proven, they're an integer and group element respectively. The point I'm trying to make here is that if you cannot state the problem, surely you can't prove it. – 2017-01-31
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    I am not proving $2n$ and $a^2$, I am proving that both of those have order $n$. Hence, look at the steps above to show you how to prove an element $a$ has an order $n$ – 2017-01-31

2 Answers 2

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If $a$ has order $2n$ then $(a^2)^n=a^{2n}=e$ and so the order $m $ of $a^2$ divides $n $. In particular $m\leq n $. On other hand we have $a^{2m}=(a^2)^m=e $. So $2n $ divides $2m $ and therefore $n $ divides $m $; in particular $n\leq m $. Therefore we conclude that $m=n $.

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1)If $a$ has order $2n$, then note that $(a^2)^n = a^{2n} = e$.

2) Suppose that $(a^2)^m = e$ for some $m$. Then, $a^{2m} =e$, and because $a$ has order $2n$, it follows that $2n | 2m$. Hence, $n | m$.

Hence, the order of $a^2$ is $n$. It is important to see how to construct proofs of this kind, I hope the above gives a good example.

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    very good example.You substituted $2n$ and $a^2$ by following the steps? – 2017-01-31
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    Exactly, that's all I have done. Please tell me if I have made it look easy, because it is. – 2017-01-31
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    lol you did make it look easy. I am new to proof. Still finding my way – 2017-01-31
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    That's ok. I've been doing this for a year. We all have this phase. – 2017-01-31