A group G has exactly 8 elements or order 3

  • Thread starter Thread starter nowimpsbball
  • Start date Start date
  • Tags Tags
    Elements Group
nowimpsbball
Messages
13
Reaction score
0
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
 
Last edited:
Mathematics news on Phys.org
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?
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top