Suppose that $G$ is a group of order $30$ and has a Sylow $5$-subgroup that is not normal. Find the number of elements of order $1$, order $2$, order $3$, and order $5$. But this scenario can't happen. Why not?

Let $G$ be a group of order $30=2 \cdot 3 \cdot 5$.

The number of Sylow $2$-subgroup $n_2$ divides $15$ and has the form $n_2=2k+1$ by the Sylow theorems. Therefore $n_2=1,3,5,15$.

The number of Sylow $3$-subgroup $n_3$ divides $10$ and has the form $n_2=3k+1$ by the Sylow theorems. Therefore $n_3=1,10$.

The number of Sylow $5$-subgroup $n_5$ divides $6$ and has the form $n_5=5k+1$ by the Sylow theorems. Therefore $n_5=1,6$. However, since Sylow $5$-subgroup isn't normal $n_5 \neq 1$.

\begin{array}{|c|c|c|c|} \hline n_2 & n_3 & n_5 & number \,of \,elements & Possible \\ \hline 1 & 1 & 6 & 1+1\cdot2+6\cdot4=26 & Yes \\ \hline 1 & 10 & 6 & 1+10\cdot2+6\cdot4=45 & No \\ \hline 3 & 1 & 6 & 3+1\cdot2+6\cdot4=29 & Yes \\ \hline 3 & 10 & 6 & 3+10\cdot2+6\cdot4=47 & No \\ \hline 5 & 1 & 6 & 5+1\cdot2+6\cdot4=31 & No \\ \hline 5 & 10 & 6 & 5+10\cdot2+6\cdot4=49 & No \\ \hline 15 & 1 & 6 & 15+1\cdot2+6\cdot4=41 & No \\ \hline 15 & 10 & 6 & 15+10\cdot2+6\cdot4=59 & No \\ \hline \end{array}

Where do I go from here?

  • 1
    $\begingroup$ Note that an element in a group of prime order has that prime order itself. Therefore none of the groups overlap but for the identity element. Add that one to your counting and you'll figure out which one it is. $\endgroup$ – Marc Dec 6 '14 at 22:49
  • $\begingroup$ So for future problems, which ever case is equal to the order of the group then that is the only case. Is this correct? $\endgroup$ – Username Unknown Dec 6 '14 at 23:06
  • $\begingroup$ The third row is the group of order 30. Hence there is 1 element of order 1. There is 3 elements of order 2. There is 2 elements of order 3. And there is 24 elements of order 5. However there isn't any elements of order 30. Is this why this scenario can not happen? $\endgroup$ – Username Unknown Dec 6 '14 at 23:10
  • $\begingroup$ Exactly, it is the third row. Keep in mind though that we used that the order of the groups are all prime, so that the groups are disjunct besides the 1 element. $\endgroup$ – Marc Dec 6 '14 at 23:29

As discussed in the comments, the third row is the only possible option. Note that in this case the subgroup of order $3$ is normal. Thus the group has a quotient of order $10$. By Sylow considerations, a group of order $10$ has a normal subgroup of order $5$ and either a normal subgroup of order $2$, in which case the group is cyclic, or $5$ subgroups of order $2$, in which case there are $5$ elements of order $2$.

Suppose the quotient is cyclic. This would mean that the quotient has an element of order $10$, which would imply that the group of order $30$ has an element either of order $10$ or order $30$, but as all of the non-identity elements have been enumerated and have prime order this is not the case.

If the quotient is not cyclic, it cannot have 5 subgroups of order $2$ because the group of order 30 only has three elements whose orders are even. Since these are the only two options for the quotient, there cannot be 6 Sylow 5-subgroups in a group of order 30, or in other words the Sylow 5-subgroup is normal.


My way: you deduced that for such a $G$ it must be $n_5=6$. These six subgroups are cyclic of order $5$ (every group of prime order is necessarely cyclic). Let $K_1,\dots,K_6$ be these $5$-Sylow. It must be $K_i\cap K_j=1$ for $i\neq j$ and every element different from the identity must has order 5. Hence $G$ contains exactly $6\cdot(5-1)=24$ elements of order $5$.

Again by your Sylow computation $n_3=1,10$. If was $n_3=10$ the by the same argument as above we'd have $10\cdot(3-1)=20$ elements of order $3$. So $G$ would contain more than $30$ elements. Hence $n_3=1$. So $G$ contains exactly $2$ elements of order $3$.

So till now we have $|G|=2\cdot3\cdot5$ hence it could contain only elements of order $2,3$ or $5$. And by Cauchy theorem it must contain at least one element of order $2,3$ and $5$.

Moreover we counted (togheter with the identity element which is the only element of order $1$) $24+2+1=27$ elements in $G$. The three remaining elements must be of order $2$, there is no other way (and from this we deduce $n_2=3$).

Summing up, we have that if $G$ is a group of order $30$ with a $5$-Sylow not normal, it must be:

$\bullet$ $n_2=3,\;\;n_3=1,\;\;n_5=6$;

$\bullet$ $G$ has exactly $1$ element of order $1$, $3$ elements of order $2$, $2$ elements of order $3$ and $24$ elements of order $5$.


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