Are Cyclic Groups with the Same Order Isomorphic?

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Homework Statement


Just want to make something clear. Are all cyclic groups that have the same number of elements isomorphic to each other.

The Attempt at a Solution


I think yes because theirs is a one-to-one correspondence and the groups are cyclic which means they have generators.
 
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Yes, all cyclic groups of order n are isomorphic. You can define the isomorphism itself by mapping the generator of one group to the generator of the other group.
 
ok thanks for your answer, can we have non-cyclic groups be isomorphic to each other.
 
Does an isomorphism between groups require that the groups be cyclic? No. Take the additive group of real numbers and the multiplicative group of positive reals. They are isomorphic via x \mapsto e^x.

However, remember that if G and G' are isomorphic groups, then G is cyclic if and only if G' is.
 
ok thanks for your answer
 
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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