Is Z8 Isomorphic to Z4 x Z2?

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SUMMARY

The discussion confirms that \(\mathbb{Z}_{8}\) is not isomorphic to \(\mathbb{Z}_{4} \times \mathbb{Z}_{2}\). This conclusion is based on the fact that the greatest common divisor (gcd) of 4 and 2 is not equal to 1, which violates the condition for isomorphism between cyclic groups. Additionally, \(\mathbb{Z}_{8}\) contains an element of order 8, while \(\mathbb{Z}_{4} \times \mathbb{Z}_{2}\) lacks such an element, further establishing their non-isomorphic nature.

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Homework Statement


Show that \mathbb{Z}_{8} is not isomorphic to \mathbb{Z}_{4}\times\mathbb{Z}_{2}.


Homework Equations


\mathbb{Z}_{mn}\cong\mathbb{Z}_{m}\times\mathbb{Z}_{n}\iff \gcd(m,n)=1


The Attempt at a Solution


I would say that since \gcd(4,2)\neq1, they are not isomorphic.

There must be something that I am misunderstanding though since \mathbb{Z}_{4} is not isomorphic \mathbb{Z}_{2}\times\mathbb{Z}_{2}...
 
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Wow...\gcd(2,2)=2...awful. Sorry for wasting space.
 
shows that Z8 has an element of order 8, and Z4 x Z2 does not.
 

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