# Is there a simple group of order $105$?

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.

• 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?
– 6666
Sep 9 '14 at 5:43
• @Joseph Every conjugate of a $p$-Sylow subgroup is a $p$-Sylow subgroup. A subgroup is normal when it has no conjugates besides itself. Sep 9 '14 at 6:03
• @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. Sep 9 '14 at 6:05
• @Joseph: Yes, sorry for for misleading remark. You-sir-33433 has given a better reason. 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 $H$.