By Sylow theorem, we see the number of the $7$-sylow subgroup is $n_7$. Then $n_7=1$ (mod $7$) and $n_7$ divides $15$; thus $n_7=15$, but why do we have $6\cdot 15=90$ elements of order $7$? And just by using $n_7,n_5,n_3$, how to answer the following question:

Is there a simple group of order $105$?

Thank you.


Two distinct $7$-Sylow subgroups intersect at the identity element (think about Langrange's Theorem). Hence, if there are $15$ distinct $7$-Sylow subgroups, it follows that, in addition to the identity element, there are at least $15\cdot 6 = 90$ elements of order $7$. But then we only have $15=105-90$ elements to account for. You can now argue that either $n_{3}=1$ or $n_{5}=1$.

Now, whenever $n_{p}=1$, this means there exists exactly one $p$-Sylow subgroup. Using Sylow's Second Theorem, this unique $p$-Sylow subgroup must be normal. As a result, the underlying group cannot be simple.

EDIT: As you-sir-33433 remarks, the normality of $p$-Sylow subgroup simply follows from the fact that $H$ and $gHg^{-1}$ have same cardinality for a subgroup $H\subset G$, and for any $g\in G$. One does not need Sylow's second theorem for this.

  • $\begingroup$ Sorry, I have a question about why unique p-Sylow subgroup must be normal, by Sylow's Second Theorem, we can only get to any two p-Sylow subgroup p1,p2 we have a number a from G such that ap1a^-1=p2, but it doesn't mean, to any a from G we have ap1a^-1 equals to another p-Sylow subgroup.So why it's normal? $\endgroup$
    – 6666
    Sep 9 '14 at 5:43
  • $\begingroup$ @Joseph Every conjugate of a $p$-Sylow subgroup is a $p$-Sylow subgroup. A subgroup is normal when it has no conjugates besides itself. $\endgroup$
    – Slade
    Sep 9 '14 at 6:03
  • $\begingroup$ @Joseph By the way, Sylow's Second Theorem is not required for this. It is required to show the converse: that a normal $p$-Sylow subgroup is unique. $\endgroup$
    – Slade
    Sep 9 '14 at 6:05
  • $\begingroup$ @Joseph: Yes, sorry for for misleading remark. You-sir-33433 has given a better reason. $\endgroup$
    – Prism
    Sep 9 '14 at 15:32

No, there does not exist a simple group of order $105$. You can view this link (see A1) to see why.

In case one does not want to navigate out of Math SE, here is what their proof says:

The number $n_3$ of Sylow $3$-subgroups of $G$ must be congruent to $1 \bmod 3$ and divide $35$. Thus $n_3 = 1$ or $7$. Similarly the number $n_5$ of Sylow $5$-subgroups is congruent to $1 \bmod 5$ and divides $21$, and so is either $1$ or $21$. If either $n_3$ or $n_5$ is $1$, the corresponding Sylow subgroup is normal and $G$ is not simple. Suppose now that $n_3=7$ and $n_5=21$. Since all the Sylow subgroups are of prime order, any two which are distinct must intersect in the identity. So if $n_3=7$ and $n_5=21$, we get $7 \cdot 2 = 14$ elements of order $3$, and $21 \cdot 4 = 84$ elements of order $5$. Since $84+14=98$, there are only $7$ elements remaining, and they must form a unique Sylow $7$-subgroup which is then normal, and $G$ is not simple.


Alternatively you may use the following lemme:

Lemma: Let $G$ be a simple group and let $H< G$ such that $[G:H]=n$. Then $G\hookrightarrow A_n$.

Why we can use here. In fact, if $|G|<\infty$ and $H< G$ such that $$[G:H]=n$$ provided $|G|\nmid n!/2$ so $G$ would be not simple. Now think of using this lemma by considering a suitable subgroup $H$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.