8
$\begingroup$

Let $G$ be a group of order $150$. I must show that it has a normal subgroup of order $25$. The hint says to show that is has a normal subgroup of order $5$ or $25$.

Now from Sylow, I know that the number $n_5$ of Sylow-$5$ subgroups (which each have $25$ elements) must be either $1$ or $6$, since $n_5$ must also divide $6$. Now clearly if $n_5=1$ we are done, so I can assume that $n_5=6$. My problem is I haven't figured out how to use this information (I am going for a contradiction, just based on what the problem asks me to prove). I know by Cauchy's theorem I can get a subgroup of order $5$, but I don't see immediately if it must be normal.

Any direction I should try to be moving?

$\endgroup$
  • $\begingroup$ Hint: Usually, one resorts to element counting at this stage. Perhaps you could assume that two of the Sylow-5 subgroups don't intersect nontrivially. $\endgroup$ – Michael Burr Jul 13 '15 at 21:55
9
$\begingroup$

As you say, by Sylow's theorem, either $n_5=1$ or $n_5=6$. If $n_5=1$, we have found a normal subgroup of $25$, so we are done. So let's assume that $n_5=6$. We will exhibit a normal subgroup of order $5$ as follows. Let $P$ and $Q$ be distinct Sylow $5$-subgroups. Using the fact that $|PQ| = \frac{|P|\cdot |Q|}{|P\cap Q|}$, it is easy to see that $|P\cap Q| = 1$ is impossible! So we conclude that the subgroup $T=P\cap Q$ has order $5$.

We claim that $T$ is normal in $G$. It is a fact if $H$ is a proper subgroup of a $p$-group $L$, then $H$ is properly contained in its normalizer $N_{L}(H)$ (see here for proof). Since $P$ and $Q$ are $p$-groups, we can apply this result. Hence, $T\subsetneq N_{P}(T)$ and $T\subsetneq N_{Q}( T)$. So $N_{P}(T) = P$ and $N_{Q}(T)=Q$. This immediately gives $P\subseteq N_{G}(T)$ and $Q\subseteq N_{G}(T)$, which implies $PQ \subseteq N_{G}(T)$. Again using the formula above, $PQ=\{p \cdot q: p\in P, q\in Q\}$ (as a subset) has cardinality $125$. By order considerations, we get that $N_{G}(T)=150$, so $N_{G}(T)=G$ and $T$ is a normal subgroup of order $5$.

This proves the hint. Now let's finally show that $G$ has a normal subgroup of order $25$. So let $T$ be the normal subgroup of order $5$ constructed above. Then the quotient group $G/T$ has order $150/5=30$. Now every group of order $30$ contains a normal Sylow $5$-subgroup (see this MSE thread). Let $H$ be a normal subgroup of order $5$ in $G/T$, and consider the canonical map $\pi: G\to G/T$. Then $\pi^{-1}(H)$ is normal subgroup of $G$. By correspondence theorem, $[G: \pi^{-1}(H)]=[G/T: H]=6$, so $\pi^{-1}(H)$ is the desired normal subgroup of index $6$, i.e. desired subgroup of order $25$.

(This shows that the case $n_5=6$ is actually impossible, and there is only one subgroup of order $25$, which is the $5$-Sylow subgroup)

$\endgroup$
3
$\begingroup$

There are easier solutions, especially if your primary goal is to show the group is not simple. (See the paragraph starting (C) for a really simple argument.)

(A) For non-simplicity: You already know by Sylow that if the group is simple it must have $6$ $5$-Sylow subgroups. If this is the case, then it acts by conjugation on these $6$ subgroups both transitively (by Sylow) and faithfully (since it's simple), and so embeds monomorphically as a subgroup of $S_6$. As $150$ does not divide $720$ (the order of $S_6$), this is impossible. So no group of order $150$ is simple.

(B) Furthermore, doing just a little more work, we can also use this homomorphism to $S_6$ to show the $5$-Sylow of any group of order $150$ is normal. The image of the homomorphism must not be divisible by $25$ (since $25$ does not divide $720$) and the image must be divisible by $6$ (to be transitive on $6$ elements). Thus, the image has order $6$ or $30$. In either event, you can see by elementary means (and probably already know) that the image has a normal $5$-Sylow, and, since the kernel is a $5$-group, the pre-image of that in $G$ is a normal $5$-Sylow in $G$.

(The group of order $30$ has a normal $5$-Sylow for example, because it has a normal subgroup of index $2$--since any element of order $2$ acts by odd permutations by translation--$15$ $2$-cycles--, so just take the elements that act by even permutations--and this normal subgroup of order $15$ both contains all $5$-Sylows and can have only one $5$-Sylow.)

Alternatively, you can do without the last paragraph-and-a-half by showing (pretty easy) that $S_6$ has no subgroup of order $30$ so the image must have order $6$.

(C) But the simplest proof of all goes like this: $G$ must have an index $2$ subgroup, since in its action on itself by multiplication, an order $2$ element maps to an odd permutation (a product of $75$ transpositions), so the elements that map to even permutations are an index $2$ subgroup $H$, which is normal and has order $75$. By Sylow's theorem, $H$ has a unique $5$-Sylow which is thus characteristic in $H$ and hence normal in $G$.

(D) Of course, if you know Philip Hall's theorem, and you know that not simple for order $150$ implies solvable, then (A) is enough since you have immediately that the $5$-Sylow is normal in any solvable group of order $150$ since the number of $5$-Sylows in a solvable group is a product of prime powers each individually congruent to $1$ mod $5$.

This embedding in $S_n$ idea works for a bunch of similar problems as well. In more complex cases, one can sometimes make more sophisticated use of this embedding in a symmetric group to make inferences about the structure of the given group; for example, if in a simple group the number of $p$-Sylows of order $p^2$ is fewer than $p^2$, then the embedding shows they must be elementary abelian rather than cyclic. This is actually kind of a big deal--if a simple group has an element of order $p^n$ all of its subgroups must have index at least $p^n$, which is an enormous constraint on the structure of the group, especially if $p$ is large. This is why so often Sylow subgroups of simple groups are elementary abelian or extra-special.

$\endgroup$
0
$\begingroup$

Just to state the simplest answer alone so people see it: Since $|G|=2m$ with $m$ odd, $G$ has an index $2$ subgroup $H$ of order $75$. By Sylow, $H$ has a unique 5-Sylow $P$ of order $25$. Since $P$ is characteristic in $H$, which is normal in $G$ (because index $2$), $P$ is normal in $G$, so it is the unique $5$-Sylow of $G$.

$\endgroup$

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.