1 maximal subgroup -> prime order

In summary: Thus $|G| = p^{r}$ for some prime $p$ and $r$.In summary, the flaw in the reasoning is that it does not follow that every element of G is an element of M. By considering a specific group where the conclusion is obviously wrong, we can see that this statement is not true. Therefore, we can conclude that if a finite group G has only one maximal subgroup M, then |G| is the power of a prime.
  • #1
boboYO
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0
'Prove that if a finite group G has only one maximal subgroup M, then |G| is the power of a prime'


I've somehow deduced that no finite group has only one maximal subgroup, and I'm having trouble seeing where I went wrong.
This is what I have:

Let [tex]H_1[/tex] be a subgroup of G. Either [tex]H_1[/tex] is maximal and equal to M or it is not maximal and there is a [tex]H_2[/tex] such that [tex]H_1<H_2<G[/tex] (using < to mean proper subgroup). Apply the same argument to [tex]H_2[/tex] and we get an ascending chain of subgroups. Since G is finite the process must end eventually (when we reach M).

Thus every subgroup of G is a subgroup of M and hence every element of G is an element of M, and G=M, a contradiction.

So would someone like to point out the flaw in the above reasoning? thanks.
 
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  • #2
boboYO said:
So would someone like to point out the flaw in the above reasoning? thanks.
Try substituting for G a particular group where the conclusion is obviously wrong. Do you know any groups with only one maximal subgroup?








The most obvious ones to me to try are the two-element group, or maybe the cyclic group on 4 elements.
 
  • #3
Thanks Hurkyl. That should've been the first thing to do. For anyone interested, the problem is with the last sentence. It does not follow that every element of G is an element of M: the elements of G that generate G do not have to be in M.
 
  • #4
I know that this post is old, but since it is top result when i google this problem, I think it would be good if i give further detail of solution.

Let $a \in G \setminus M$. Then $a$ generate $G$ which mean $G$ is cyclic. Suppose $|G| = p_{1}^{r_{1}}...p_{n}^{r_{n}}$ where $p_{i}$ is prime. So $a^{p_{i}}$ and $a^{p_{j}}$ generated a different maximal subgroup which is contradiction.
 

Related to 1 maximal subgroup -> prime order

1. What is a maximal subgroup?

A maximal subgroup is a subgroup within a larger group that is not a proper subset of any other subgroup. In other words, it is the largest subgroup that still maintains the same structure and properties as the original group.

2. How do you determine if a subgroup is maximal?

To determine if a subgroup is maximal, you can use the Lagrange's theorem. If the order of the subgroup evenly divides the order of the original group, then it is a maximal subgroup.

3. What is the significance of a subgroup having prime order?

A subgroup with prime order is significant because it allows for more efficient computations and algorithms when working with the group. It also has implications in cryptography and number theory.

4. Can a maximal subgroup have an order that is not prime?

Yes, a maximal subgroup can have an order that is not prime. However, if the maximal subgroup has an order that is not prime, then it must have a prime power order. This means that the order can be written as a prime number raised to a positive integer power.

5. How does the concept of maximal subgroups relate to group theory?

The concept of maximal subgroups is an important aspect of group theory. It allows for the classification and study of different types of groups, as well as the understanding of their structures and properties. Maximal subgroups also play a key role in the study of symmetry and other mathematical concepts.

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