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I have a homework question, the subject matter of which the class I'm in has covered terribly:

"Given a group $A$, which has an order of $4$ and has an element of order $4$, prove that $A$ is isomorphic to $\Bbb{Z}_{4}$. "

I feel like this question should be easy, so I am also wondering if there are any resources which will help me sharpen my theoretical math skills.

edit: I have since realized that what I was asking was more specifically, how to begin the problem. Intuitively, I knew that $|A| = | \Bbb{Z}_{4} |$ , but in order to proceed with the problem I needed clues as to what the set $G$ looks like, namely $G = \{ e=1= a^{4}, a^{1}, a^{2}, a^{3} \} $ under multiplication $mod$ $a^{4}$.

Thomas
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    The resources are texts about abstract algebra, which you could read. Let $x$ be the element of order $4$. Then $A={e,x,x^2,x^3}$, which is isomorphic to $C_4$. – Dietrich Burde Jan 29 '18 at 21:51
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    Dietrich... Also known as $\mathbb Z_4 = {0, 1, 2, 3}$ under addition, where both 1, and 3, are of order 4. – amWhy Jan 29 '18 at 21:58
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    Hint: Any group of order $k$ which has an element of order $k$ is cyclic, and isomorphic to any other order-$k$ cyclic group. – Joffan Jan 29 '18 at 23:06

2 Answers2

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Since $a$ has order $4$, the elements $1=a^0$, $a$, $a^2$ and $a^3$ are distinct. Since $|A|=4$, we conclude $$ A=\{a^0,a,a^2,a^3\}=\langle a\rangle $$ which means $A$ is cyclic.

egreg
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The key idea is that if a group $A$ of order $4$ has an element $a$ of order 4, then we know that $a$ generates $A$, i.e., then $$A= \langle a \rangle = \{a^0, a^1, a^2, a^3\}.$$

In other words, $A$ is cyclic, such that $a^4 = a^0, a^5 = a^1,$ etc.

And all cyclic groups of order $4$ are isomorphic to the group $\mathbb Z_4$ under addition, mod $4$.

There are exactly two groups of order $4$, up to isomorphism:

  1. The Klein-4 group: $\{e, a, b, c\}$ such that $ab = c, ac = b$, and $bc = a$, and in which all elements $a, b, c$ are of order two: $a^2 = b^2 = c^2 = e.$

  2. $\mathbb Z_4 = \{0, 1, 2, 3\}$ under addition, modulo 4. This group has one element of order 1, which is $0$, the identity, one element of order $2$ which happens to be $2$, and two elements of order $4$: $1, 3$. Note $\mathbb Z_4$ is a cyclic group of order $4$ which is generated by either $1$ or $3$. Since the group operation is addition, modulo $4$, $\{a^0, a^1, a^2, a^3\}$, when using the generator $a = 1$, can be written: $$\{o\cdot 1, 1\cdot 1, 2\cdot 1, 3\cdot 1\} = \{0, 1, 2, 3\}$$

Since you are given that $A$ has at least one element of order $4$, $A$ cannot be isomorphic the the Klein-4 group, in which, except for the identity e, all elements are of order 2. That leaves us to conclude that $A \cong \mathbb Z_4$.

amWhy
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