### Simple problem1

If G is a finite group such that 3 does not divide o(G), and $a^3b^3=(ab)^3$ for all a,b in G, then G is abelian.javascript:;

Hint. Consider the homomorphism $f:G\to G:x\mapsto x^3$, of which the kernel is trivial (for 3 doen't divide |G|). So f is a monomorphism i.e. injective and as G is finite, f is bijective.

Now see that $ab.ab.ab = aaa.bbb\;\;$ which means:

(1)
$$ba.ba =a^2b^2$$

again, $(xyx^{-1})^3=x^3y^3x^{-3}\Rightarrow xy^3x^{-1}=x^3y^3x^{-3}\Rightarrow y^3x^2=x^2y^3$ i.e., squares commute with qubes. But as f is surjective as well, every element is a qube, and squares commute with everything, so $ba.ba=a^2b^2=b^2a^2$, from 1. This is enough.

### Group of order pq

Let $|G|=pq$, where $p>q$ are primes.
Now, the Sylow p-subgroup, say P, is normal. And G/P is of order q, hence is abelian. This means, $G'\subseteq P$.

If a Sylow q-subgroup Q is normal, then similarly, $G'\subseteq Q$. This means, $G'\subseteq P\cap Q=1$. Hence G is abelian.

### A is in the center of G

Let $|G|=n$, and $A\trianglelefteq G$ where $|A|=p$, prime and $\gcd(p-1,n)=1$.

Then, consider the map $G\to \Aut(A):g\mapsto \iota_g:x\mapsto gxg^{-1}$. The order of $\iota_g$ divides $|G|=n$, and as $\iota_g\in\Aut(A)\cong \mathbb Z_p^\times$, its order divides $(p-1)$. So, the order divides $\gcd(p-1,n)=1$, hence each such $\iota_g$ is identity. This proves that A is in the center of G.

### Group of odd order

Problem In a group of odd order no element other than the identity is conjugate to its inverse.

Solution. Assume that the group is G and its order is (2s-1) where s is positive integer.
If an element $x$ is conjugate with its inverse, then there is $g\in G$ with $gxg^{-1}=x^{-1}$.

Now take the subgroup $H=\langle x\rangle$ and see that it is invariant under $\hat g\colon z\mapsto gzg^{-1}$ because $\hat g(x)=gx^ig^{-1}=x^{-i}$. The map $G\to G:g\mapsto \hat g$ is a homomorphism i.e., $\hat a\hat b=\widehat{ab}$. Observe that in its restriction to $H$, $(\hat g)^2=\widehat {g^2}=id_H$.

Now, as the order of the group is odd, we see that $\hat g=\widehat {g^{2s}}=(\widehat {g^2})^s=id_H$, whence $x^{-1}=\hat g(x)=x$ so that $x^2=e_G$. But $x=x^{2s}=(x^2)^s=e_G$.$\qquad \Box$

### Herstein's famous problem

Problem . If $G$ is a finite abelian group then if it has two subgroups of order m and n, then it must have a subgroup of order lcm(m,n).

While this follows easily from the observation that in case of finite abelian groups, the converse of Lagrange's theorem holds, the result can also be proved from preliminary concepts.

Lemma. Given m,n coprime, a finite abelian group G has a subgroup of order mn if and only if it has subgroups of order m and order n.

Proof. If H and K are subgroups with their orders m and n respectively, then H+k has order mn.

If A is a subgroup of order mn, then $nA$ has order m, etc..

Now, if $m=p_1^{\alpha_1}\cdots p_k^{\alpha_k},\; n=p_1^{\beta_1}\cdots p_k^{\beta_k}$, and so, by the lemma, G has subgroups of order $p_i^\max[\alpha_i,\beta_i]$, for each $i$, and thus, a subgroup of order $lcm(m,n)$.

### Group of order 2n

Let G be a finite group of order 2n, and half its elements be of order=2, the rest form a subgroup of order n.

Call the last subgroup H. If n were even, there is an element of order 2 in H. So,
Lemma 1. $n$ is odd.

Now, there are only two left cosets of H in G i.e., [G:H]=2. Hence H is normal in G.

Lemma 2. $H\trianglelefteq G$.

Let $G/H=\{H,bH\}$. Call $C_2=\{1,b\}$. This is a group since b is of order 2. Here we observe one thing : the map $x\mapsto bxb$ is an automorphism of G, and if $a\in H$, then $ab$ can't be in H, since then b would be in H. In that way, $ab$ is of order 2. So, $ab\cdot ab=1$, and thus, $bab=a^{-1}$. This means,

Lemma 3. $H$ is stable under the automorphism $\iota_b:G\to G:x\to bxb^{-1}$, and the restriction of this map to $H$ is inversion. Hence, inversion is an automorphism of $H$.

In view of Lemma 3, we can say that given $x,y\in H$, we have that

(2)
\begin{eqnarray} (xy)^{-1}&=&x^{-1}y^{-1}\\ {\rm or,}\quad xy&=&yx.\\ \end{eqnarray}

Hence,
Theorem 4. $H$ is abelian.

Note:
Now, $G=C_2H=HC_2$, and $C_2\cap H=\{1\}$. Hence $G\cong H\rtimes_\theta C_2$ where

(3)
\begin{align} \theta:C_2\to\Aut(H):b\mapsto\theta_b:a\mapsto bab^{-1}. \end{align}

### Three quarter

Theorem 1. Let $G$ be a finite group, and suppose that an automorphism $T$ of $G$ sends more than three-quarters of the elements of $G$ onto their inverses. Then $T\left(x\right) = x^{ - 1}$ for all $x\in G$, and that $G$ is abelian.

Proof.

1. Define $H: = \{ x \colon T(x) = x^{ - 1} \}$ .

2. Choose $a\in H$.

3. Define $H_1:=\{x\in H\colon ax\in H\}=H\cap a^{-1} H$.

4. Lemma: if $A\subseteq G$, then $x\mapsto yx$ defines a one-to-one correspondence between $A$ and $yA$. Hence $|A|=|yA|,\forall y\in G$.

5. $|G|\ge |H\cup a^{-1} H|=|H|+|a^{-1} H|-|H\cap a^{-1}H|$, whence

(4)
\begin{align} |H\cap a^{-1}H|\ge |H|+|a^{-1}H|-|G|>\frac{3}{4}|G|+\frac{3}{4}|G|-|G|=\frac{|G|}{2}. \end{align}

6. Given $x\in H_1,$ we see that $x^{-1} a^{-1}=(ax)^{-1}=T(ax)=T(a)T(x)=a^{-1} x^{-1}$ , whence $ax=xa$. Hence $H_1\subseteq C(a)$, the centralizer of $a$.

7. As $H_1$ has more than half the elements of $G$, the index of $C(a)(\supseteq H_1)$ would be less than 2, so that $C(a)=G$, i.e., $(\forall x\in G)(ax=xa)$.

8. But the last assertion of 7 holds for any $a$ in $H$, for the choice of $a$ was arbitrary. So, $(\forall y\in H)(\forall x\in G)(yx=xy)$, so that $H\subseteq Z(G)$, the centre of $G$. But

(5)
\begin{eqnarray} \frac{|G|}{2}<|H_1|\le|H|\le|Z(G)|, \end{eqnarray}

so that the index of $Z(G)$ in $G$ is less than 2, and $Z(G)=G$.

### Solvable pqr

Lemma A finite group $G$ of order $pqr$, where $p<q<r$ are primes, has at least one of the Sylow subgroups normal in it.

Proof. If not, then for each of these primes, the number of the corresponding Sylow subgroups would be greater than 1.

Number of r-subgroups is 1 or $pq$, hence there are exactly $pq(r-1)=pqr-pq$ elements of order $r$.

Number of $q$-subgroups would divide $pr$, but is >1, hence is >$p$. So the number of elements of order $q$ is at least $r(q-1)=qr-r$.

The number of $p$-subgroups is not one, and would divide $qr$, so it is at least $q$. The number of elements of order $p$ is thus, at least $q(p-1)$. The total number of elements of $G$ is at least

(6)
\begin{eqnarray} pqr-pq+qr-r+pq-q+1&=&pqr+(r-1)(q-1)\\ &>&pqr,\;\mbox{ since $q,r$ being prime $>1$,} \end{eqnarray}

which is impossible since the group is of order $pqr$.$\hfill\Box$

The Sylow subgroups of $G$ are solvable, for they are abelian; and the quotient group of $G$ by the normal Sylow subgroup is of order $\alpha\beta$ where $\alpha<\beta$ are primes. This has its $\beta$-subgroup normal and the quotient by it, abelian (cyclic, actually). So it is solvable itself. Hence the group $G$ is solvable, too. We summarise the result:

THEOREM Any finite group of order $pqr$ where $p<q<r$ are primes, is solvable.

### normal subgroup whose index and order are coprime

Let $G$ be a finite group, $N\trianglelefteq G$ and let $[G:N]$ be coprime to $|N|$.

Now, if we have $|N|$ in the form $p_1^{\alpha_1}\cdots p_k^{\alpha_k}$, then

(7)
\begin{align} |G|=p_1^{\alpha_1}\cdots p_k^{\alpha_k}\cdot m \end{align}

where $m$ is prime to each $p_i$. As each $p_i$-Sylow subgroup of $N$ is also a $p_i$-Sylow subgroup of $G$, and as any other $p_i$-Sylow subgroup of $G$ would be a conjugate of that contained in $N$, and as $N$ is normal, it contains ALL the $p_i$-Sylow subgroups of $G$.

If there were any "other" subgroup $A$ of $G$ with the order $|N|=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$. Consider the subgroup $N\cap A$. For each $p_i$,\; $A$ contains some $p_i$-Sylow subgroup, but that Sylow one is contained in $N$ also, by the preceding discussion. So this is contained in $N\cap A$. Hence for each $p_i$, its highest power, i.e., $p_i^{\alpha_i}$ divides $|N\cap A|$. It means $p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ divides $|N\cap A|$. But this is impossible since $N\cap A$ is contained strictly in $N$. So we obtain:

Theorem If $G$ is a finite group with $N$ normal in it, such that $[G:N]$ and $|N|$ are coprime to each other, then there is no other subgroup of $G$ of order $|N|$.

### Edit me to add new problem

page revision: 20, last edited: 02 Sep 2010 11:10