What is the significance of a group with no proper subgroups?

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    2015
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SUMMARY

A group \( G \) with no proper subgroups is necessarily a trivial group, meaning it contains only the identity element. This conclusion is established through the definition of groups and the properties of subgroups. The only group that satisfies this condition is the group containing just the identity element, denoted as \( G = \{ e \} \). This result is significant in group theory as it highlights the fundamental nature of groups and their structure.

PREREQUISITES
  • Understanding of group theory concepts, particularly the definition of groups and subgroups.
  • Familiarity with the identity element in group structures.
  • Basic knowledge of mathematical proofs and logical reasoning.
  • Experience with abstract algebra terminology.
NEXT STEPS
  • Study the properties of trivial groups in group theory.
  • Explore the classification of groups based on their subgroup structures.
  • Learn about the implications of groups with minimal elements in abstract algebra.
  • Investigate examples of groups with various subgroup configurations.
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Mathematicians, students of abstract algebra, and anyone interested in the foundational aspects of group theory will benefit from this discussion.

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Here is this week's POTW:

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Suppose $G$ is a group with no proper subgroups. What can be said about $G?$ Prove your statements.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to johng for his correct solution to this week's POTW, which follows:

$G$ is finite and cyclic of order 1 or cyclic of order a prime.
Let $E=\langle 1\rangle$ be the identity subgroup of $G$ and suppose $G\neq E$. Let $x\in G$ with $x\neq 1$. Then $\langle x\rangle=G$ and $x$ has finite order, for otherwise $\langle x^2\rangle$ is a proper subgroup of $G$. Suppose $x$ has order $n$ with $n$ composite, say $n=mq$. Then $\langle x^m\rangle$ has order $q$ and so would be a proper subgroup of $G$. Hence $n$ is prime. Thus $G=\langle x\rangle$ is cyclic of prime order $n$.
 

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