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I just wanted to practice my proofs and my understanding of Isomorphic so I decided to prove the following if I am wrong or need a better argument for anything please feel free to let me know so I can correct it for a better understanding

Theorem: Let $\phi$ be a group isomorphism of $G$ onto $G_1$ then

1) $\phi^{-1}$ is an isormophism from $G_1$ to $G$

2) $G$ is abelian if and only if $G_1$ is abelian

3) $G$ is cyclic if and only if $G_1$ is cyclic

4)If $K$ is a subgroup of G, then $ \phi(K)$ is a subgroup of $G_1$

Proof(s) 1)Assume $P:G\rightarrow G_1$. $P$ is $1-1$, onto, and operation preserving i.e a bijection therefore $P^{-1}$ must be a bijection as well meaning $P^{-1}(a*b)= P^{-1}(a)*P^{-1}(b)$

2)(Forward direction) Assume $G$ is abelian, then $ab=ba$ for $a,b\in G$. so $P(ab)=P(ba)$, so $P(a)P(b)=P(b)P(a)$. So for $P:G\rightarrow G_1$, $P(a)*P(b)=P(b)+P(a)$. (Backward direction) Assuming $G_1$ is abelian then the same argument can be said but instead using $c,d$ within $G_1$ and using $P^{-1}$

3)(Forward Direction): Assume $G$ is cyclic, so there is some element a within $G$ that generates the entire group $G$. So every event in $G$ has a form $a^n$ for some a within $G$ and some $n$ within $\mathbb{Z}$. Since $P(a^n)=(P(a))^n$, everything in $G_1$ has a form $(P(a))^n$, so $G_1$ is generated by $<P(a)>$ which makes $G_1$ a cyclic group.

(backward direction)Assume $G_1$ is cyclic, so $<P(a)>$ generates $G_1$, so there is a $<P^{-1}(P(a)>$ that generates $G$.

4)Assume $K$ is a subgroup of $G$. To prove $P(K)$ is a subgroup of $G_1$ Ill use to two step test. First $P(K)$ is non empty because $e$ within $k$ so $P(e_1)$ is within $P(K)$

a)Suppose a within $P(K)$ so $a^{-1}= P(U^{-1})$ for some $U$ in $P(K)$, $e_1=P(e)= P(U*U^{-1})$ so $P(U^{-1})$ but $U^{-1}$ is within $K$ so $a^{-1} P(U^{-1})$ is within $P(K)$

b)Pick any two elements of $P(K)$, $P(t),P(s)\in P(K)$. Product $P(t)P(s)=P(st)$ and $st$ is within $k$ since $k$ is a subgroup so $P(s)P(t)$ is within $P(K)$

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  • $\begingroup$ You should probably take the time to $\LaTeX$ this properly... $\endgroup$ Nov 7 '13 at 3:19
  • $\begingroup$ This is basically fine. Perhaps you could give more detail in 1). You do have a bijection of sets, but you need to say more about why the group operations correspond. $\endgroup$ Mar 19 '15 at 15:45
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I guess you are asking for suggestions / comments for your proof. Here is one thing that I noticed to simplify your argument in the backward direction for 2 and 3. Suppose for both 2 and 3 we have the forward directions. We can simply say by 1, there is an isomorphism $\phi^{-1}$ from $G$ to $G_1$. Apply the forward directions to this isomorphism and we get the backward directions.

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