A group G has exactly 8 elements or order 3

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A group G has exactly 8 elements of order 3, leading to the question of how many subgroups of order 3 exist within G. The prime decomposition of 8 is 2^3, suggesting multiple ways to form factors, but clarification is needed on the interpretation of these factors in relation to subgroups. Each subgroup of order 3 contains elements of order 3, prompting inquiries about the number of such elements and the intersection of distinct subgroups. The discussion highlights the need for precise terminology regarding the elements and their orders in group theory. Understanding the structure of G is crucial for determining the number of subgroups of order 3.
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A group G has exactly 8 elements of order 3 (Unanswered as of 1/31)

How many subgroups of order 3 does G have?

So we have 8 elements, its prime decomposition is 8=2^3. The number of different ways to get factors is how many subgroups, at least that is what I interpret from my notes...so there are 2^3, 2*2^2, and 2*2*2, so three different ways to get factors, am I doing this right?

Thanks
 
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Do you mean G has 8 elements OR order 3, or do you mean G has 8 elements OF order 3.
 
d_leet said:
Do you mean G has 8 elements OR order 3, or do you mean G has 8 elements OF order 3.

g has 8 elements OF order 3, my bad
 
every subgroup of order three has how many elements of order three?

and how many common elements of order three do two distinct subgroups of order three have?
 
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