A map $\phi:G\to G’$ is called a group homomorphism if
\[\phi(ab)=\phi(a)\phi(b)\]
for any elements $a, b\in G$.
Proof.
Let $e, e’$ be the identity elements of $G, G’$, respectively.
First we claim that
\[\phi(e)=e’.\]
In fact, we have
\begin{align*}
\phi(e)&=\phi(ee)=\phi(e)\phi(e) \tag{*}
\end{align*}
since $\phi$ is a group homomorphism.
Thus, multiplying by $\phi(e)^{-1}$ on the left, we obtain
\begin{align*}
&e’=\phi(e)^{-1}\phi(e)\\
&=\phi(e)^{-1}\phi(e)\phi(e) && \text{by (*)}\\
&=e’\phi(e)=\phi(e).
\end{align*}
Hence the claim is proved.
Then we have
\begin{align*}
&e’=\phi(e) && \text{by claim}\\
&=\phi(gg^{-1})\\
&=\phi(g)\phi(g^{-1}) && \text{since $\phi$ is a group homomorphism}.
\end{align*}
It follows that we have
\begin{align*}
\phi(g)^{-1}&=\phi(g)^{-1}e’\\
&=\phi(g)^{-1}\phi(g)\phi(g^{-1})\\
&=e’\phi(g^{-1})\\
&=\phi(g^{-1}).
\end{align*}
This completes the proof.
A Group Homomorphism and an Abelian Group
Let $G$ be a group. Define a map $f:G \to G$ by sending each element $g \in G$ to its inverse $g^{-1} \in G$.
Show that $G$ is an abelian group if and only if the map $f: G\to G$ is a group homomorphism.
Proof.
$(\implies)$ If $G$ is an abelian group, then $f$ […]
Pullback Group of Two Group Homomorphisms into a Group
Let $G_1, G_1$, and $H$ be groups. Let $f_1: G_1 \to H$ and $f_2: G_2 \to H$ be group homomorphisms.
Define the subset $M$ of $G_1 \times G_2$ to be
\[M=\{(a_1, a_2) \in G_1\times G_2 \mid f_1(a_1)=f_2(a_2)\}.\]
Prove that $M$ is a subgroup of $G_1 \times G_2$.
[…]
Basic Properties of Characteristic Groups
Definition (automorphism).
An isomorphism from a group $G$ to itself is called an automorphism of $G$.
The set of all automorphism is denoted by $\Aut(G)$.
Definition (characteristic subgroup).
A subgroup $H$ of a group $G$ is called characteristic in $G$ if for any $\phi […]
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Show that a group $G$ is cyclic if and only if there exists a surjective group homomorphism from the additive group $\Z$ of integers to the group $G$.
Proof.
$(\implies)$: If $G$ is cyclic, then there exists a surjective homomorhpism from $\Z$
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Let $G$ be a group and define a map $f:G\to G$ by $f(a)=a^2$ for each $a\in G$.
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Proof.
$(\implies)$ If $G$ is an abelian group, then $f$ is a homomorphism.
Suppose that […]
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Let $\Z$ be the additive group of integers. Let $f: \Z \to \Z$ be a group homomorphism.
Then show that there exists an integer $a$ such that
\[f(n)=an\]
for any integer $n$.
Hint.
Let us first recall the definition of a group homomorphism.
A group homomorphism from a […]
Abelian Groups and Surjective Group Homomorphism
Let $G, G'$ be groups. Suppose that we have a surjective group homomorphism $f:G\to G'$.
Show that if $G$ is an abelian group, then so is $G'$.
Definitions.
Recall the relevant definitions.
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Let $G$ and $H$ be groups and let $f:G \to K$ be a group homomorphism. Prove that the homomorphism $f$ is injective if and only if the kernel is trivial, that is, $\ker(f)=\{e\}$, where $e$ is the identity element of $G$.
Definitions/Hint.
We recall several […]
[…] Suppose that we have [phi(a)=phi(b)] for some elements $a, bin G$. Then the properties of the homomorphism $phi$ imply that begin{align*} phi(a)phi(b)^{-1}=e’\ phi(a)phi(b^{-1})=e’\ phi(ab^{-1})=e’. end{align*} In the second step, we used the fact $phi(b)^{-1}=phi(b^{-1})$, which is proved in the post “Group Homomorphism Sends the Inverse Element to the Inverse Element“. […]
[…] In the second step, we used the fact $f(g_2^{-1})=f(g_2)^{-1}$, which is proved in the post “Group Homomorphism Sends the Inverse Element to the Inverse Element“. […]
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[…] Suppose that we have [phi(a)=phi(b)] for some elements $a, bin G$. Then the properties of the homomorphism $phi$ imply that begin{align*} phi(a)phi(b)^{-1}=e’\ phi(a)phi(b^{-1})=e’\ phi(ab^{-1})=e’. end{align*} In the second step, we used the fact $phi(b)^{-1}=phi(b^{-1})$, which is proved in the post “Group Homomorphism Sends the Inverse Element to the Inverse Element“. […]
[…] In the second step, we used the fact $f(g_2^{-1})=f(g_2)^{-1}$, which is proved in the post “Group Homomorphism Sends the Inverse Element to the Inverse Element“. […]