The multiplicative group $ \mathbb {{F}^{\times}_{7}}$ is isomorphic to a subgroup of multiplicative group $ \mathbb {{F}^{\times}_{31}}$

Question

The multiplicative group $ \mathbb {{F}^{\times}_{7}}$ is isomorphic to a subgroup of the multiplicative group $ \mathbb {{F}^{\times}_{31}}$.

Can anyone tell me this statement is true or false. Please explain your answer. I am new to algebra.

Answer 1

Yes, it is true. The group $\mathbb{Z}_7^\times$ is a cyclic group of order $6$ (it is generated by $3$). And the group $\mathbb{Z}_{31}^\times$ has an element of order $6$, which is $6$.

Answer 2

José Carols Santos' answer is fine; but I'll try to be a bit more systematic in mine for other people reading this question (and who know maybe a bit more algebra ? )

If $K$ is a field and $G\subset K^\times$ is a finite subgroup of the multiplicative group of $K$; then it is cyclic. Indeed, let $n= exp(G)$ (the exponent of $G$, i.e. the lowest common multiple of the orders of the elements of $G$). By definition, for all $x\in G, x^n =1$, so that $G$ is included in the set of roots of $X^n-1$. But this set has size at most $n$, so $|G| \leq n$. However, by Lagrange's theorem, $exp(G)\mid |G|$ so $n\leq |G|$, so that $n=|G|$.

By standard methods, $G$ being abelian implies it has an element of order $exp(G)$. By what came before, this is an element of order $|G|$: in other words, $G$ is cyclic.

Now this applies to $\mathbb{F}_q^\times$ which is indeed a finite subgroup of the multiplicative group of a field; hence it is cyclic. Its order is $q-1$.

Now a cyclic group of order $n$ has precisely one subgroup of order $d$ for each $d\mid n$, and this subgroup is (clearly) cyclic. This implies that $\mathbb{F}_q^\times$ is isomorphic to a subgroup of $\mathbb{F}_r^\times$ if and only if $q-1\mid r-1$.

In particular, in your situation, since $7-1 = 6, 31-1= 30$ and $6\mid 30$, we find that $\mathbb{F}_7^\times$ is isomorphic to a subgroup of $\mathbb{F}_{31}^\times$. A way to show that this is true is, as José explained in his answer, to pick the element $6$ which has (multiplicative) order $6$ modulo $31$.

Answer 3

The other two answers are excellent, but they take the top-down approach. For a beginner like you, I recommend a bottom-up approach via hand computation.

First, write down the powers of $3$ modulo $7$: $3^0=1,3^1=3,3^2=9\equiv2,3^3\equiv3\cdot2=6,3^4\equiv3\cdot6\equiv4,3^5\equiv5,3^6\equiv1$. So you see you’ve hit all the nonzero residues modulo $7$.

Now do the same for congruence modulo $31$. You’ll find that $3^4\equiv19$, $3^{10}\equiv25$, all the way up to $3^{15}\equiv30\equiv-1$. To do all this, you did only $14$ multiplications in all, never getting a product bigger than $90$. Now complete the list, making use, if you choose, of the fact that $3^{15}\equiv-1$. What have you shown yourself? That $\Bbb F_{31}^*$ is cyclic, generated by $3$, and that if you take its fifth power, namely $26$, you’ll be holding in your hand a congruence class that generates a subgroup of order six. Yay!