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I'm trying to solve this question for a while and I think I'm missing something.
if you can please help me without using "Cauchy's theorem".
My try
By Lagrange's theorem we know that for each $g \in G : o(g)||G| \Rightarrow o(g)|22$
So we have only $4$ options : $o(g)=1 , o(g)=2 , o(g)=11 , o(g)=22$
Let's cancel the option $o(g)=1$ because then $g=e$ and it won't help us.
if $o(g)=22 \rightarrow G$ is cyclic so we can pick $o(g^2)=11$ and $o(g^{11})=2$ so we are done.

I'm stuck in this phase of if $o(g)=2$ or $o(g)=11$
I tried proving by contradiction that it can't be that exist more than $1$ elemet of order $2$ or $11$
I know that the amount of elements of order $2$ is odd (since $22$ is even).
I know that $2$ and $11$ is prime.

I noticed I didn't use the fact that $G$ is abelian so I guess I need to use it somehow yet I don't know how.
Any clues would help.

Have a nice day and thank you!

Roach87
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  • I don't understand the downvotes. Roach has made a clear attempt at the problem. – Shaun Mar 21 '21 at 13:31
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    @Shaun I agree, there is an attempt. I am only a bit confused about the source of this problem. It seems copied wrongly, or not? It should be clarified. Why the command about not using Cauchy's Theorem? – Dietrich Burde Mar 21 '21 at 14:00

1 Answers1

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For an odd prime $p$, an abelian group of order $2p$ is cyclic, so we have $G\cong C_{2p}$. A cyclic group of order $n$ has a unique subgroup of order $d$ for each divisor $d\mid n$.

So there is a unique subgroup of order $2$ resp. $11$ for $G$ with $|G|=2p=2\cdot 11$. However, the number of elements of order $d$ in $G$ is given by $\phi(d)$. We have $\phi(2)=1$ and $\phi(11)=10$, so that there is a unique element of order $2$ in $G$, but certainly not a unique element of order $11$.

References:

Looking for a simple proof that groups of order $2p$ are up to isomorphism $\mathbb{Z}_{2p}$ and $D_{p}$ .

Each finite cyclic group of order $n$ contains unique subgroup of order $d$ where $d\mid n$

Dietrich Burde
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