### 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)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 $x=x^{2s}=(x^2)^s=e_G$.$\qquad \Box$

For other methods, see this page

### 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)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

### 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)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

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)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)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 $|N|$.*