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Prove that a group $G$ of order $375$ has a subgroup of order $15.$

My thought:

$5|O(G)\implies G$ has a subgroup $H$ of order $5.$

$3|O(G)\implies G$ has a subgroup $K$ of order $3.$

$|H\cap K|=(e)$ and so $|HK|=\dfrac{|H|.|K|}{|H\cap K|}=15$

Now if I can show $HK=KH$ then we're done. But I don't know if it holds at all!

Sriti Mallick
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2 Answers2

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By the Sylow theorems, the group certainly has a normal Sylow $5$-subgroup, and either has a normal sylow Sylow $3$-subgroup, or else has 25 Sylow $3$-subgroups.

Let $F$ mean a $5$-subgroup and $T$ mean a Sylow $3$-subgroup. If $F$ and $T$ are normal, then they centralize each other, and your argument works (why?)

If instead $T$ isn't normal, there are $25$ Sylow 3's. But the number of Sylow 3's is the index of the normalizer $N_G(T)$. What does this say about its order?

rschwieb
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Hint: $n_3$ is $1$ or $25$ by Sylow theorems. If $n_3 = 1$, then you can show $HK = KH$, where $H$ and $K$ are as you defined. If $n_3 = 25$, check the size of the normalizer $N_G(K)$.