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Prove that $\Bbb{Z}/8\Bbb{Z}$ and $\Bbb{Z}/4\Bbb{Z}\times \Bbb{Z}/2\Bbb{Z}$ are not isomorphic.

Proof: We know that if $y\in\Bbb{Z}/8\Bbb{Z}$, then $\max_{y\in\Bbb{Z}/8\Bbb{Z}}\left\{ \text{ord}(y) \right\}=8$, but if $x\in\Bbb{Z}/4\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}$, then $\max_{x\in\Bbb{Z}/4\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}}\left\{ \text{ord}(x) \right\}=4$. Therefore $\Bbb{Z}/4\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}\not\cong\Bbb{Z}/8\Bbb{Z}$.

Is this a valid proof? Ifso can I do the same thing for $\Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}$, any tips/hints?

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yes. that's true and enugh. –  Mj125 Dec 2 '12 at 18:30
    
@Mj125 Thanks for the quick reply. –  Onur Dec 2 '12 at 18:33

1 Answer 1

up vote 6 down vote accepted

Your observation that $\Bbb Z/4\Bbb Z\times\Bbb Z/2\Bbb Z$ has no element of order $8$ is correct and shows that the group cannot be isomorphic to $\Bbb Z/8\Bbb Z$.

A similar argument uses the observation that in $\Bbb Z/4\Bbb Z\times\Bbb Z/2\Bbb Z$ one can find more than one subgroup consisting of $2$ elements. This shows that the group cannot be cyclic as a cyclic group of order $n$ has a unique subgroup of order $d$ for each divisor $d$ of its order.

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@Jl90: In fact $\Bbb Z/4\Bbb Z\times\Bbb Z/2\Bbb Z$ is abelian but another group is cyclic. This is what you can above. Nice explanation. +1 –  Babak S. Dec 2 '12 at 19:06

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