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Right now, I am in the process of showing that a group $G$ of order $35$ must be cyclic, for the purpose of then showing that there is a unique group of order $35$, namely $\mathbb{Z}_{35}$.

To that effect, I was interested in applying an argument seen in the answer to the linked question given by Ben. Ben is attempting to prove that $G$ such that $|G|=35$ must be cyclic by using the fact that the Third Sylow Theorem tells us that the number of subgroups of order $5$ must be equal to $1 \mod 5$ and the number of subgroups of order $7$ must be equal to $1 \mod 7$. In order to prove that there can be only one such subgroup of order $7$, he assumes that there are instead, $8$ such subgroups and attempts to arrive at a contradiction. He states that "if there are $8$ subgroups of order $7$, then each one contains $6$ elements plus the identity so there must be $6 \cdot 8 = 48$ distinct elements in the group", which since the order of $G$ is only $35$ must be a contradiction. My question is, how does he know that for any two of those $p$-subgroups, call them $H$ and $K$ that $H \cap K = \{e\}$? It's possible that is a Sylow-related result I am unaware of or missed, or it could be possible that that was an erroneous assumption on Ben's part.

Now, you may ask, "why don't you just ask Ben"? I would, were it not for the fact that Ben has only $35$ reputation, has not been seen on the site since 2015, and therefore, is very unlikely to answer me should I do so.

I thank you ahead of time.

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The subgroups are of prime order (not prime power order, which is worth noting). So if $H$ and $K$ contain some element $g$ in common other than the identity, then they both also contain all powers of $g$. So you can't have one non-identity element in common without having at $6$ non-identity elements in common. And at that point, $H=K$.


Alternatively, the intersection of two subgroups is a group. So $H\cap K$ is a subgroup of $H$. But $H$ has order $7$, so its subgroups are only $H$ itself and the trivial group. So either $H=H\cap K$ (which implies $H=K$ since both have order $7$) or $H\cap K=\{e\}$ (meaning the two subgroups only have one element in common.)

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  • $\begingroup$ in which case, supposing the prime in question is $7$, it would degenerate to the case where there is only $1$ subgroup of order $7$? $\endgroup$ – ALannister Mar 16 '17 at 2:59
  • $\begingroup$ just to make sure I'm actually making sense here, it does follow that if I can show that $G$ where $|G|=35$ is cyclic, and if any cyclic group of order $35$ is isomorphic to $\mathbb{Z}_{35}$, that I will have succeeded in showing that there is a unique group of order $35$ up to isomorphism, right? I just want to check, because I'm very sleep-deprived and don't trust my brain right now. $\endgroup$ – ALannister Mar 16 '17 at 3:39
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    $\begingroup$ Yes. If a generic "two" subgroups of order $7$ are demonstrated to be the same subgroup, then there is only one subgroup of order $7$. And yes to your second comment. $\endgroup$ – alex.jordan Mar 16 '17 at 3:49
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Let $H$ and $K$ be groups of order $p$ where $H\neq K$.
Then $H\cap K\leq H$.
By Lagrange's Theorem, $|H\cap K|=1$ or $p$.
This means that $H\cap K=H$ or $1$.
Suppose $H\cap K=H$. Then $H\leq K$.
Since $|H|=|K|$, this means that $H=K$ which is a contradiction.
So we conclude that $H\cap K=1$

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  • $\begingroup$ just to make sure I'm actually making sense here, it does follow that if I can show that $G$ where $|G|=35$ is cyclic, and if any cyclic group of order $35$ is isomorphic to $\mathbb{Z}_{35}$, that I will have succeeded in showing that there is a unique group of order $35$ up to isomorphism, right? I just want to check, because I'm very sleep-deprived and don't trust my brain right now. $\endgroup$ – ALannister Mar 16 '17 at 3:39
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    $\begingroup$ @ALannister Yes you are right! $\endgroup$ – Alan Wang Mar 16 '17 at 3:41

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