Is Z8 Isomorphic to Z4 x Z2?

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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.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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