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Let $G$ be a group of order $120$ with a normal subgroup $N$ of order 5. Let $H$ be any subgroup of $G$ of order $15$. Prove $N$ is a subgroup of $H$.

I have a proof (see below), but I am wondering: Is there a simpler proof? In particular, something that doesn't use that $HN \simeq H \times N$.

Proof: Since $N$ is normal, $HN$ is a subgroup $G$. By Lagrange, since $N$ has prime order it must be cyclic and every non-identity element is a generator. By way of contradiction, suppose $H$ does not contain $N$. Then $H \cap N = \{e\}$. From this, it follows that $HN \simeq H \times N$. Thus $|HN| = |H||N| = 15 \cdot 5 = 75$. By Lagrange, $75 \mid 120$. Contradiction.

Shaun
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JasonJones
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    You don't need $HN\cong H\times N$. You only need that $HN$ is a subgroup. By standard counting argument $|HN|= |H||N|/|H\cap N|$. Since $|H\cap N|=1$ or $5$, if it were $1$ you get the contradiction you want just as you did, and so $|H\cap N|=5$. This means $H\cap N=N$. – Arturo Magidin Sep 19 '23 at 20:17
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    If you know Sylow, then it is immediate, since $5$ is the largest power of $5$ that divides $120$. – Arturo Magidin Sep 19 '23 at 20:23
  • And you should know Sylow (or at least his theorems). – Dietrich Burde Sep 19 '23 at 20:29
  • "By Lagrange, since $N$ has prime order it must be cyclic and every non-identity element is a generator.": I don't think you need this part. – Kan't Sep 19 '23 at 20:39
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    I mean, the primality of $5$ suffices – Kan't Sep 19 '23 at 20:46
  • @citadel Good point. For future readers: https://math.stackexchange.com/questions/106163/show-that-every-group-of-prime-order-is-cyclic – JasonJones Sep 20 '23 at 17:27
  • @ArturoMagidin Can you please sketch the counting argument? – JasonJones Sep 20 '23 at 17:34
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    It's a standard argument: Map $(h,k)$ to $hk$. $(h,k)$ maps to the same element as $(h',k')$ if and only if there exists $x\in H\cap K$ such that $h'=hx$ and $k'=x^{-1}k$. That means every element in $HK$ is the image of $|H\cap K|$ elements from $H\times K$, so $|H|,|K| = |H\times K| = |HK|,|H\cap K|$. For finite quantities, this gives $|HK|=|H||K|/|H\cap K|$. This holds whenever $H$ and $K$ are subgroups, regardless of whether $HK$ is a subgroup or not. It should be in every basic group theory book. – Arturo Magidin Sep 20 '23 at 17:41
  • @ArturoMagidin Thanks. I'm trying to work out the Sylow argument, would you mind taking a look?: $n_5(H)$ = number of 5-Sylow subgroups of $H$. $n_5(G)$ analogous. By Sylow's third theorem, $n_5(H) \equiv n_5(G) \equiv 1$ mod $5$. Sylow 5-subgroups of $H$ have order 5. Same for G. If $N$ is not a subgroup of $H$, then $n_5(G) \geq n_5(H) \geq 6$. Then what? – JasonJones Sep 20 '23 at 18:13
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    @JasonJones: By Sylow's 2nd Theorem any two Sylow subgroups of $G$ are conjugate. That means that $G$ has a normal $5$-Sylow subgroup if and only if it has a unique $5$-Sylow subgroup. So $G$ has a unique subgroup of order $5$. Every subgroup of order $15$ has a subgroup of order $5$, which is also a subgroup of order $5$ of $G$. – Arturo Magidin Sep 20 '23 at 18:18
  • @ArturoMagidin Not a criticism. But isn't the counting argument at least as hard as proof that $HN \simeq H \times N$? I guess you don't need to define the product group $H \times N$ to understand it though. And it is certainly more general. Thanks for your help. – JasonJones Sep 20 '23 at 18:19
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    @JasonJones: Point the first: The counting argument is a standard argument that appears very early on in any standard group theory book, so one would expect you to know it well ahead of knowing what an internal direct product of two subgroups is. Point the Second: you asked for a proof "that doesn't use that $HN\cong H\times N$". I gave you a proof (in fact, two, one basic and one advanced) that does not use that. Isn't that exactly what you asked for? – Arturo Magidin Sep 20 '23 at 18:24
  • @ArturoMagidin All of that is fair. Like I said, not a criticism. Thank you. – JasonJones Sep 20 '23 at 18:30
  • Your solution seems fine to me. In the sense that it is more or less equivalent to my gut reaction solution (and Arturo's variant). The answer by testaccount (+1) is probably the simplest, but it relies on the uniqueness (immediate from Sylow, among other things). – Jyrki Lahtonen Sep 22 '23 at 11:05

1 Answers1

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Let $G$ be a group of order $|G| = 120$ with a normal subgroup $N$ of order $|N| = 5$.

Note that since $\gcd(|G/N|, |N|) = 1$, it follows that $N$ is the only subgroup of order $5$ in $G$.

Now let $H \leq G$ be a subgroup of order $15$. Because $H$ contains a subgroup of order $5$ (for example by Cauchy's theorem, or by Sylow's theorem), conclude that $N \leq H$.

testaccount
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