Show that a subgroup of order $3$ is normal in a group of order $15$

Question

I've been trying to show that any group of order $15$ is cyclic and the only missing part in my proof is to show that the subgroup with $3$ elements (which we know exists by Cauchy's theorem) is actually normal.

I've shown the normality of the subgroup of order $5$ by its uniqueness but such a trick does not work with $3$ because $3$ is too small. I know by the 3rd Sylow theorem that $N\equiv 1 \mod 3$ where $N$ is the number of subgroups of order $3$ and since $N\mid 15$ it must be $1$. So this would imply that the subgroup is unique hence normal.

But is there a way to show this result without resorting to Sylow?

Answer 1

If we know that in a group $G$ of order $15$ a subgroup of order $5$ is normal, then it follows easily that $G$ is a cyclic group.

Indeed let $a$ and $b$ be elements of the group $G$ of order $3$ and $5$ respectively. Since the subgroup $\langle b\rangle$ is normal in $G$, we have $$ a^{-1}ba=b^k, $$ where $k$ is one of the numbers $\{1,2,3,4\}$. Next $$ b=a^{-3}ba^3=a^{-2}b^ka^2=(a^{-2}ba^2)^k=\ldots=b^{k^3}. $$ Hence $k^3\equiv1\pmod5$. It follows that $k=1$. The rest is clear.

Answer 2

If $G$ is abelian, then you are done. Otherwise, the only possible class equation is$^\dagger$: $$15=1+3k+5l$$ a contradiction, because then the normal (as you found out already) subgroup of order $5$ cannot be built up as union of conjugacy classes.

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$^\dagger$For $x\in G\setminus Z(G)$, we get (strictly): $Z(G)<C_G(x)<G$, so necessarily $|Z(G)|\ne3,5$.

Answer 3

The proof can directly follows from the application of sylows theorem which states that if order of G is pq where p and q are distinct primes and p<q.if p doesnot divide q-1 then G is cyclic Now |G|=15=3×5 and 3 doesnot divide (5-1)=4 so by above theorem G is cyclic hence abelian and every subgroup of abelian group group is normal. So that subgroup of order 3 is normal(there will be a subgroup of order 3 by cauchy's theorem for finite abelian group)

Answer 4

This is an OVERKILL to the problem using character theory.
We denote the subgroup of order $3$ as $H$. We shall prove that $H\subseteq\text{Core}_G(H)$. If we denote the trivial character of $H$ as $1_H$, it holds that $$\text{Core}_G(H)=\bigcap_{g\in G}H^g=\bigcap_{g\in G}\ker(1_H)^g=\ker(1_H)^G.$$ Where $(1_H)^G$ is the induced character of $1_H$. Now, recall that $\ker(1_H)^G$ is the intersection of the kernels of the irreducible constituents of $(1_H)^G$.
Let $\psi$ be an irreducible constituent of $(1_H)^G$. It suffices to prove that $H\subseteq\ker\psi$.

If $\psi$ is linear then, by Frobenius reciprocity, $$0\not=[(1_H)^G,\psi]=[1_H,\psi_H],$$ and hence $1_H=\psi_H$ since $\psi_H$ is linear (it is irreducible). In this case $H\subseteq\ker\psi$.

If $\psi$ is not linear then $\psi(1)=3$ (since $\psi(1)||G|$ and $\psi(1)^2\leq|G|$). Note that $\ker\psi\vartriangleleft G$. If $H\not\subseteq\ker\psi$ then it would hold that $H\ker\psi=G$. Let's see that this is not possible.

Assume that $H\ker\psi=G$ and consider $\Psi$ a representation affording $\psi$. Then $\mathbb C^n$ is a $\mathbb C G$-module via $vg=v\Psi(g)$ which affords $\psi$. Notice that this $\mathbb C G$-module is simple since $\psi$ is irreducible. Therefore $(\mathbb C^n)_H$ is a $\mathbb C H$-module affording $\psi_H$. If this module wasn't simple then it would exists a nontrivial $\mathbb C H$-submodule of $(\mathbb C^n)_H$, say $W$. But this implies that for each $w\in W$ and $k\in\ker\psi$ it holds that \begin{equation*} wk=w\Psi(k)=w\Psi(1)=w\in W. \end{equation*} This makes $W$ a nontrivial $\mathbb C G$-submodule of our $\mathbb C^n$ since $G=(\ker\psi)H$, which is a contradiction. Thus $\psi_H$ is irreducible.
But if this is the case, it must hold that $9=\psi_H(1)^2\leq |H|=3$, which is a contradiction.

Hence $H$ is contained in the kernel of all the irreducible constituents of $(1_H)^G$, or equivalently, H is contained in $\text{Core}_G(H)$, and we are done.