Some basic problems of group theory

Question

1.Prove the every group G of order 4 is isomorphic to either Z4 or 4-group V,that is {1 (1,2)(3,4) (1,3)(2,4) (1,4)(2,3)}

2.If G is a group with $|G|\leq 5$ then G is abelian.

I have learned independently the chapter 1 of the book An introduction to group theory by Joseph Rotman . And can understand it ,can solve many exercises in it.I can understand the proofs in chapter 2 , however I totally don't have any intuition of the exercises. Does any suggestion?

Answer 1

Chandru1 Answered 2 pretty adequately. I'll focus on part 1 of your question.

Take $x\in G$ not the identity. By Lagrange's theorem the order of this element is either $2$ or $4$. If it is order four, you can map it to the generator $1$ of $\mathbb{Z}_4$. It is not hard to establish that this generates an isomorphism.

If $x$ has order $2$, you can look to the other elements. If none of these elements are of order $4$, you have that every square in this group is the identity. It is not hard to establish the required isomorphism to $V$ in this case.

Answer 2

Note:

• If $|G|=4$, then $G$ is either isomorphic $\mathbb{Z}_{4}$ or $\mathbb{Z}_{2} \times \mathbb{Z}_{2}$. Infact, if $|G|=p^{2}$, then either $G$ is isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_{p} \times \mathbb{Z}_{p}$. For a proof of this you can see Dummit and Foote's Abstract Algebra book. ( I don't know where it's exactly given.)

• If $|G|=1$, then it's abelian, if $|G|=2$, it's cyclic, since $2$ is prime so it's abelian, if $|G|=3$, it's cyclic so abelian, if $|G|=4$, by above it's abelian, if $|G|=5$, it's cyclic so abelian.

Answer 3

Here is an alternate proof for (2), which is also a hint for 1:

We first prove the following:

(*) If any element in $G$ has order $2$, then $G$ is abelian.

Proof for (*): Let $x,y \in G$. Then

$$xxyy=ee=e=(xy)^2=xyxy \,.$$

Multiplying to the left by $x^{-1}$ and to the right by $y^{-1}$ yields

$$xy=yx \,.$$

Now (2) follows imediatelly from (*): By Lagrange Theorem any group of prime order is cyclic, thus abelian. For the case $|G| =4$, either there is an element of order $4$, or every element has order $2$; in both cases we get $G$ is abelian.