Group of order 100 with no element of order 4?

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1. Homework Statement [/]
Is there a group G with order 100 such that it has no element of order 4? How would one go about proving the existence of such a group?



2. Homework Equations [/]
For every prime divisor p of a group, there exists an element with order p.



The Attempt at a Solution

 
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Do you know about direct products? If so, you should be able to construct such a group pretty easily.
 
Yes I do have some knowledge of direct products. Could I construct a group as such: \mathbb{Z}_{5}\times\mathbb{Z}_{5}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}?
 
kpoltorak said:
Yes I do have some knowledge of direct products. Could I construct a group as such: \mathbb{Z}_{5}\times\mathbb{Z}_{5}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}?

That's the one I had in mind.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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